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Combinations and Permutations What's the Difference? In English we use the word "combination" loosely, without thinking if the order of things is important. In other words: "My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad. "The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2. So, in Mathematics we use more precise language: If the order doesn't matter, it is a Combination. If the order does matter it is a Permutation. A Permutation is an ordered Combination.

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Page 1: Combinations and permutations

Combinations and Permutations

What's the Difference?

In English we use the word "combination" loosely, without thinking if the order of things is important. In other words:

"My fruit salad is a combination of apples, grapes and bananas" We don't care what order the fruits are in, they could also be "bananas, grapes and apples" or "grapes, apples and bananas", its the same fruit salad.

   "The combination to the safe was 472". Now we do care about the order. "724" would not work, nor would "247". It has to be exactly 4-7-2.

So, in Mathematics we use more precise language:

If the order doesn't matter, it is a Combination.If the order does matter it is a Permutation.

A Permutation is an ordered Combination.

Permutation : Permutation means arrangement of things. The word arrangement is used, if the order of things is considered.

Combination: Combination means selection of things. The word selection is used, when the order of things has no importance.

Definition:

Permutation:          An arrangement is called a Permutation. It is the rearrangement of objects or symbols into distinguishable sequences. When we set things in order, we say we have made an arrangement. When we change the order, we say we have changed the 

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arrangement. So each of the arrangement that can be made by taking some or all of a number of things is known as Permutation.

Combination:          A Combination is a selection of some or all of a number of different objects. It is an un-ordered collection of unique sizes.In a permutation the order of occurence of the objects or the arrangement is important but in combination the order of occurence of the objects is not important.

Formula:

Permutation = nPr = n! / (n-r)! Combination = nCr = nPr / r! where,              n, r are non negative integers and 0≤ r≤n.              r is the size of each permutation.              n is the size of the set from which elements are permuted.              ! is the factorial operator.

Example:     Suppose we have to form a number of consisting of three digits using the digits 1,2,3,4, To form this number the digits have to be arranged. Different numbers will get formed depending upon the order in which we arrange the digits. This is an example of Permutation.

Now suppose that we have to make a team of 11 players out of 20 players, This is an example of combination, because the order of players in the team will not result in a change in the team. No matter in which order we list out the players the team will remain the same! For a different team to be formed at least one player will have to be changed.

Now let us look at two fundamental principles of counting:

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Addition rule : If an experiment can be performed in ‘n’ ways, & another experiment can be performed in ‘m’ ways then either of the two experiments can be performed in (m+n) ways. This rule can be extended to any finite number of experiments.

Example:       Suppose there are 3 doors in a room, 2 on one side and 1 on other side. A man wants to go out from the room. Obviously he has ‘3’ options for it. He can come out by door ‘A’ or door ‘B’ or door ’C’.

 

Multiplication Rule : If a work can be done in m ways, another work can be done in ‘n’ ways, then both of the operations can be performed in m x n ways. It can be extended to any finite number of operations.

Example.:      Suppose a man wants to cross-out a room, which has 2 doors on one side and 1 door on other site. He has  2 x 1  = 2 ways for it.

Factorial n : The product of first ‘n’ natural numbers is denoted by n!.

            n!   = n(n-1) (n-2) ………………..3.2.1.

            Ex.       5! = 5 x 4 x 3 x 2 x 1 =120

            Note       0!     =  1

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            Proof   n! =n, (n-1)!

            Or           (n-1)! = [n x (n-1)!]/n = n! /n                     

            Putting n = 1, we have

            O!  = 1!/1 = 1

            

Permutation

Number of permutations of ‘n’ different things taken ‘r’ at a time is given by:-

nPr       =            n!/(n-r)!

Proof:     Say we have ‘n’ different things a1, a2……, an.

Clearly the first place can be filled up in ‘n’ ways. Number of things left after filling-up the first place = n-1

So the second-place can be filled-up in (n-1) ways. Now number of things left after filling-up the first and second places = n - 2

Now the third place can be filled-up in (n-2) ways.

Thus number of ways of filling-up first-place = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place = n-2

Number of ways of filling-up r-th place = n – (r-1) = n-r+1

By multiplication – rule of counting, total no. of ways of filling up, first, second --  rth-place together :-

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n (n-1) (n-2) ------------ (n-r+1)

Hence:nPr       = n (n-1)(n-2) --------------(n-r+1)

= [n(n-1)(n-2)----------(n-r+1)] [(n-r)(n-r-1)-----3.2.1.] / [(n-r)(n-r-

1)] ----3.2.1

nPr = n!/(n-r)!

Number of permutations of ‘n’ different things taken all at a time is given by:-

nPn               =         n!

Proof   : Now we have ‘n’ objects, and n-places.

Number of ways of filling-up first-place  = n

Number of ways of filling-up second-place = n-1

Number of ways of filling-up third-place  = n-2

Number of ways of filling-up r-th place, i.e. last place =1

Number of ways of filling-up first, second, --- n th place= n (n-1) (n-2) ------ 2.1.

nPn =  n!

Concept. 

We have   nPr  =     n!/n-r

Putting r = n, we have :-

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nPr  =   n! / (n-r)

But   nPn  =  n!

Clearly it is possible, only when  n!  = 1

Hence it is proof that     0! = 1

Note : Factorial of negative-number is not defined. The expression  –3! has no meaning.

Examples

Q. How many different signals can be made by 5 flags from 8-flags of different colours?

Ans.    Number of ways taking 5 flags out of 8-flage  = 8P5

=   8!/(8-5)!       

=  8 x 7 x 6 x 5 x 4 = 6720

Q. How many words can be made by using the letters of the word “SIMPLETON” taken all at a time?

Ans.   There are ‘9’ different letters of the word “SIMPLETON”

Number of Permutations taking all the letters at a time  = 9P9

=  9!    = 362880.

Number of permutations of n-thing, taken all at a time, in which ‘P’ are of one type, ‘q’ of them are of second-type, ‘r’ of them are of third-type, and rest are all different is given by :-

  n!/(p! x q! x

r!)                                                                                               

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Example: In how many ways can the letters of the word “Pre-University” be arranged?

13!/(2! X 2! X 2!)

Number of permutations of n-things, taken ‘r’ at a time when each thing can be repeated r-times is given by = nr.

Proof.   

Number of ways of filling-up first –place   = n

Since repetition is allowed, so

Number of ways of filling-up second-place  = n

Number of ways of filling-up third-place             

Number of ways of filling-up r-th place  = n

Hence total number of ways in which first, second ----r th,

places can be filled-up

 =  n x n x n ------------- r factors.

=   nr

Example:  John has 8 friends. In how many ways can he invite

one or more of them to dinner?

Ans.    John can select one or more than one of his 8 friends.

=> Required number of ways = 28 – 1= 255.

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(iv) Number of ways of selecting zero or more things from ‘n’

identical things is given by :- n+1

Example:   In how many ways, can zero or more letters be

selected form the letters AAAAA?

Ans. Number of ways of :

         Selecting zero 'A's   =  1

         Selecting one 'A's   =   1

         Selecting two 'A's     = 1

         Selecting three 'A's  =  1

         Selecting four 'A's   =   1

         Selecting five 'A's    =  1

=> Required number of ways  = 6         [5+1]

(V)    Number of ways of selecting one or more things from ‘p’

identical things of one type ‘q’ identical things of another

type, ‘r’ identical things of the third type and ‘n’ different

things is given by :-

 (p+1) (q+1) (r+1)2n – 1

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Example:    Find the number of different choices that can be

made from 3 apples, 4 bananas and 5 mangoes, if at least

one fruit is to be chosen.

Ans:

Number of ways of selecting apples = (3+1) = 4 ways.

Number of ways of selecting bananas = (4+1) = 5 ways.

Number of ways of selecting mangoes = (5+1) = 6 ways.

Total number of ways of selecting fruits = 4 x 5 x 6

But this includes, when no fruits i.e. zero fruits is selected

=> Number of ways of selecting at least one fruit = (4x5x6) -1  =

119

Note :- There was no fruit of a different type, hence here  n=o

 =>   2n  = 20=1

(VI)   Number of ways of selecting ‘r’ things from ‘n’ identical things is ‘1’.

Example:   In how many ways 5 balls can be selected from ‘12’ identical red balls?

Ans. The balls are identical, total number of ways of selecting 5 balls  = 1.

Example: How many numbers of four digits can be formed with digits 1, 2, 3, 4 and 5?

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Ans. Here n = 5                     [Number of digits]

And   r = 4                     [ Number of places to be filled-up]

Required number is   5P4 = 5!/1!  = 5 x 4 x 3 x 2 x 1

Example:      A child has 3 pocket and 4 coins. In how many ways can he put the coins in his pocket.

Ans.    First coin can be put in 3 ways, similarly second, third and forth coins also can be put in 3 ways.

So total number of ways = 3 x 3 x 3 x 3   = 34   = 81

So, we should really call this a "Permutation Lock"!

Permutations

There are basically two types of permutation:

1. Repetition is Allowed: such as the lock above. It could be "333". 

2. No Repetition: for example the first three people in a running race. You can't be first and second. 

 

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1. Permutations with Repetition

These are the easiest to calculate.

When you have n things to choose from ... you have n choices each time!

When choosing r of them, the permutations are:

n × n × ... (r times)

(In other words, there are n possibilities for the first choice, THEN there are n possibilites for the second choice, and so on, multplying

each time.)

Which is easier to write down using an exponent of r:

n × n × ... (r times) = nr

Example: in the lock above, there are 10 numbers to choose from (0,1,..9) and you choose 3 of them:

10 × 10 × ... (3 times) = 103 = 1,000 permutations

So, the formula is simply:

nr

where n is the number of things to choose from, and 

you choose r of them(Repetition allowed, order 

matters) 

 

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2. Permutations without Repetition

In this case, you have to reduce the number of available choices each time.

For example, what order could 16 pool balls be in?

After choosing, say, number "14" you can't choose it again.

So, your first choice would have 16 possibilites, and your next choice would then have 15 possibilities, then 14, 13, etc. And the total

permutations would be:

16 × 15 × 14 × 13 × ... = 20,922,789,888,000

But maybe you don't want to choose them all, just 3 of them, so that would be only:

16 × 15 × 14 = 3,360

In other words, there are 3,360 different ways that 3 pool balls could be selected out of 16 balls.

But how do we write that mathematically? Answer: we use the "factorial function"

The factorial function (symbol: !) just means to multiply a series of descending natural numbers. Examples:

4! = 4 × 3 × 2 × 1 = 24  7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5,040  1! = 1 

Note: it is generally agreed that 0! = 1. It may seem funny that multiplying no numbers together gets you 1, but it helps simplify a lot of equations.

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So, if you wanted to select all of the billiard balls the permutations would be:

16! = 20,922,789,888,000

But if you wanted to select just 3, then you have to stop the multiplying after 14. How do you do that? There is a neat trick ... you

divide by 13! ...

16 × 15 × 14 × 13 × 12 ...

  = 16 × 15 × 14 = 3,360

13 × 12 ...

Do you see? 16! / 13! = 16 × 15 × 14

The formula is written:

where n is the number of things to choose from, and 

you choose r of them(No repetition, order matters) 

Examples:

Our "order of 3 out of 16 pool balls example" would be:

16!

=

16!

=

20,922,789,888,000

= 3,360

(16-3)! 13! 6,227,020,800

(which is just the same as: 16 × 15 × 14 = 3,360)

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How many ways can first and second place be awarded to 10 people?

10!

=

10!

=

3,628,800

= 90

(10-2)! 8! 40,320

(which is just the same as: 10 × 9 = 90)

Notation

Instead of writing the whole formula, people use different notations such as these:

Example: P(10,2) = 90

Combinations

There are also two types of combinations (remember the order does not matter now):

1. Repetition is Allowed: such as coins in your pocket (5,5,5,10,10) 

2. No Repetition: such as lottery numbers (2,14,15,27,30,33)  

 

How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?

A84

B120

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C504

D720Answer :The number of permutations of 3 digits chosen from 10 is 10P3 = 10 × 9 × 8 = 720

Jones is the Chairman of a committee. In how many ways can a committee of 5 be chosen from 10 people given that Jones must be one of them?

A126

B252

C495

D3,024 Answer:  Jones is already chosen, so we need to choose another 4 from 9.In choosing a committee, order doesn't matter; so we need the number of combinations of 4 people chosen from 9= 9C4 = 9!/(4!)(5!)= (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1)= 3,024/24  = 126

A password consists of four different letters of the alphabet. How many different possible passwords are there?

(Using pen and paper can help you learn.)A

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426

B456,976

C14,950

D358,800Answer: The number of permutations of 4 letters chosen from 26 is 26P4 = 26 × 25 × 24 × 23 = 358,800A password consists of two letters of the alphabet followed by three digits chosen from 0 to 9. Repeats are allowed. How many different possible passwords are there?

(Don't worry if you get it wrong ... you can learn from your mistakes.)

A492,804

B650,000

C676,000

D1,757,600Answer: The number of ways of choosing the letters = 26 × 26 = 676The number of ways of choosing the digits = 10 × 10 × 10 = 1,000

So the number of possible passwords = 676 × 1,000 = 676,000

An encyclopedia has eight volumes. In how many ways can the eight volumes be replaced on the shelf?

(Use pen and paper to work out the answer.)A8

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B5,040

C40,320

D88

Answer:Imagine there are 8 spots on the shelf. Replace the volumes one by one.The first volume to be replaced could go in any one of the eight spots.The second volume to be replaced could then go in any one of the seven remaining spots.The third volume to be replaced could then go in any one of the six remaining spots.etc

So the total number of ways the eight volumes could be replaced = 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1= 40,320Assuming that any arrangement of letters forms a 'word', how many 'words' of any length can be formed from the letters of the word SQUARE?

(No repeating of letters)

(No guessing! Be sure of your answer.)A82

B720

C1,956

D9,331

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Answer: The number of one letter 'words' = 6P1 = 6The number of two letter 'words' = 6P2 = 6 × 5 = 30The number of three letter 'words' = 6P3 = 6 × 5 × 4 = 120The number of four letter 'words' = 6P4 = 6 × 5 × 4 × 3 = 360The number of five letter 'words' = 6P5 = 6 × 5 × 4 × 3 × 2 = 720The number of six letter 'words' = 6P6 = 6! = 720

So the total number of possible 'words' = 6 + 30 + 120 + 360 + 720 + 720 = 1,956

16 teams enter a competition. They are divided up into four Pools (A, B, C and D)  of four teams each.

Every team plays one match against the other teams in its Pool.

After the Pool matches are completed:• the winner of Pool A plays the second placed team of Pool B • the winner of Pool B plays the second placed team of Pool A • the winner of Pool C plays the second placed team of Pool D• the winner of Pool D plays the second placed team of Pool C

The winners of these four matches then play semi-finals, and the winners of the semi-finals play in the final.

How many matches are played altogether?

(Using pen and paper can help you learn.)A23

B31

C32

D63

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Answer:The number of matches played in each Pool = 4C2 = 4!/(2!2!) = (4 × 3)/(2 × 1) = 6

So the total number of Pool matches = 4 × 6 = 24

The winners and second placed teams play a further 4 matches.Then there are 2 semi-finals and 1 final

So the total number of matches = 24 + 4 + 2 + 1 = 31

A restaurant offers 5 choices of appetizer, 10 choices of main meal and 4 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses. Assuming all choices are available, how many different possible meals does the restaurant offer?

A329

B310

C200

D19Answer; A person who eats only an appetizer has 5 choices.A person who eats only a main meal has 10 choices.A person who eats only a dessert has 4 choices.

A person who eats an appetizer and a main meal has 5 × 10 = 50 choices.A person who eats an appetizer and a dessert has 5 × 4 = 20 choices.A person who eats a main meal and a dessert has 10 × 4 = 40 choices.

A person who eats all three courses has 5 × 10 × 4 = 200 choices

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So the total number of possible meals = 5 + 10 + 4 + 50 + 20 + 40 + 200 = 329

Question:

How many ways can 4 prizes be given away to 3 boys,

if each boy is eligible for all the prizes?

(1)256

(2)12

(3)81

(4) None of these

Correct Answer - (3)

Solution:

Any one prize can be given to any one of the 3 boys and hence there are 3 ways of distributing each prize.

Hence, the 4 prizes can be distributed in 34= 81 ways.

Question:

How many words of 4 consonants and 3 vowels can be

made from 12 consonants and 4 vowels, if all the

letters are different?

(1)16C7.7!

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(2)12C4.4C3.7!

(3)12C3.4C4

(4) 12C4 . 4C3

Correct Answer - (2)

Solution:

4 consonants out of 12 can be selected in 12C4 ways.

3 vowels can be selected in 4C3 ways.

Therefore, total number of groups each containing 4 consonants and 3 vowels = 12C4 . 4C3

Each group contains 7 letters, which can be arranging in 7! ways.

Therefore required number of words = 124 . 4C3 . 7!

Question:

In how many ways can the letters of the word

EDUCATION be rearranged so that the relative position

of the vowels and consonants remain the same as in

the word EDUCATION?

(1)9!/4

(2)9!/(4!.5!)

(3)4!.5!

(4) None of these

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Correct Answer - (3)

Solution:

The word EDUCATION is a 9 letter word, with none of the letters repeating.

The vowels occupy 3, 5, 7th and 8th position in the word and the remaining 5 positions are occupied by consonants

As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the aforementioned 4 places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th positions.

The 4 vowels can be arranged in the 3rd, 5th, 7th and 8th position in 4! Ways.

Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6th and 9th position in 5! Ways.

Hence, the total number of ways = 4! . 5!.Question:

There are 12 yes or no questions. How many ways can

these be answered?

(1)1024

(2)2048

(3) 4096

(4) 144

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Correct Answer - (3)

Solution:

Each of the questions can be answered in 2 ways (yes or no)

Therefore, no. of ways of answering 12 questions = 212

= 4096 ways.Question:

What is the value of 1.1! + 2.2! + 3!.3! + ............ n.n!,

where n! means n factorial or n(n-1)(n-2)...1

Solution:

1.1! = (2 -1).1! = 2.1! – 1.1! = 2! - 1!2.2! = (3 - 1).2! = 3.2! - 2! = 3! - 2!3.3! = (4 - 1).3! = 4.3! - 3! = 4! - 3!......n.n! = (n+1 - 1).n! = (n+1)(n!) - n! = (n+1)! - n!

Summing up all these terms, we get (n+1)! - 1!Question:

In how many ways can the letters of the word

MANAGEMENT be rearranged so that the two As do not

appear together?

Solution:

The word MANAGEMENT is a 10 letter word.

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Normally, any 10 letter word can be rearranged in 10! ways.

However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.

Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to

.

The problem requires us to find out the number of outcomes in which the two As do not appear together.

The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter. Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can

be rearranged in ways.

Therefore, the required answer in which the two As do not appear next to each other =

Total number of outcomes - the number of outcomes in which the 2 As appear together

=> ways.**Question:

There are 5 Rock songs, 6 Carnatic songs and 3 Indi

pop songs. How many different albums can be formed

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using the above repertoire if the albums should contain

at least 1 Rock song and 1 Carnatic song?

(1)15624

(2)16384

(3)6144

(4) 240

Correct Answer - (1)

Solution:

There are 2n ways of choosing ‘n’ objects. For e.g. if n = 3, then the three objects can be chosen in the following 23 ways - 3C0 ways of choosing none of the three, 3C1 ways of choosing one out of the three, 3C2 ways of choosing two out of the three and 3C3 ways of choosing all three.

In the given problem, there are 5 Rock songs. We can choose them in 25 ways. However, as the problem states that the case where you do not choose a Rock song does not exist (at least one rock song has to be selected), it can be done in 25 - 1 = 32 - 1 = 31 ways.

Similarly, the 6 Carnatic songs, choosing at least one, can be selected in 26 - 1 = 64 - 1 = 63 ways.

And the 3 Indi pop can be selected in 23 = 8 ways. Here the option of not selecting even one Indi Pop is allowed.

Therefore, the total number of combinations = 31 . 63 . 8 = 15624Question:

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How many words can be formed by re-arranging the

letters of the word ASCENT such that A and T occupy

the first and last position respectively?

(1)5!

(2)4!

(3)2!

(4) 6! / 2!

Correct Answer - (2)

Solution:

As A and T should occupy the first and last position, the first and last position can be filled in only one way. The remaining 4 positions can be filled in 4! Ways by the remaining words (S,C,E,N,T). hence by rearranging the letters of the word ASCENT we can form 1x4! = 4! words.**Question:

A team of 8 students goes on an excursion, in two cars,

of which one can seat 5 and the other only 4. In how

many ways can they travel?

Solution:

There are 8 students and the maximum capacity of the cars together is 9.

We may divide the 8 students as follows

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Case I: 5 students in the first car and 3 in the second

Or Case II: 4 students in the first car and 4 in the second

Hence, in Case I: 8 students are divided into groups of 5 and 3 in 8C3 ways.

Similarly, in Case II: 8 students are divided into two groups of 4 and 4 in 8C4 ways.

Therefore, the total number of ways in which 8 students can travel is 8C3 + 8C4 = 56 + 70 = 126.**Question:

When four fair dice are rolled simultaneously, in how

many outcomes will at least one of the dice show 3?

Solution:

When 4 dice are rolled simultaneously, there will be a total of 64 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 54 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

Question:

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How many ways can 10 letters be posted in 5 post

boxes, if each of the post boxes can take more than 10

letters?

Solution:

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the 10 letters.

i.e. 5.5.5….5 (upto 10 times)

= 510.

Here is another way to calculate it:

Including "none" as an option, there are 6 choices of appetizer, 11 choices of main meal and 5 choices of dessert.  Thus the total number of choices is 6x11x5=330. 

One of these is not a meal though (no appetizer, no main meal and no dessert), so there are 329 possible meals.

Circular permutations

There are two cases of circular-permutations:-

(a)       If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!

(b)       If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by  (n-1)!/2!

Proof(a):         

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(a)       Let’s consider that 4 persons A,B,C, and D are sitting around a round table

Shifting A, B, C, D, one position in anticlock-wise direction, we get the following agreements:-

 

Thus, we use that if 4 persons are sitting at a round table, then they can be shifted four times, but these four arrangements will be the same, because the sequence of A, B, C, D, is same. But if A, B, C, D, are sitting in a row, and they are shifted, then the four linear-arrangement will be different.

 

Hence if we have ‘4’ things, then for each circular-arrangement number of linear-arrangements =4

Similarly, if we have ‘n’ things, then for each circular – agreement, number of linear – arrangement = n.

Let the total circular arrangement  = p

Total number of linear–arrangements = n.p

Total number of linear–arrangements 

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= n. (number of circular-arrangements)

Or Number of circular-arrangements = 1 (number of linear arrangements)

 n  = 1( n!)/n

circular permutation = (n-1)!

Proof (b)   When clock-wise and anti-clock wise arrangements are not different, then observation can be made from both sides, and this will be the same. Here two permutations will be counted as one. So total permutations will be half, hence in this case.

Circular–permutations =  (n-1)!/2

Note:   Number of circular-permutations of ‘n’ different things taken ‘r’ at a time:-

(a)  If clock-wise and anti-clockwise orders are taken as different, then total number of circular-permutations  =   nPr /r

(b) If clock-wise and anti-clockwise orders are taken as not different, then total number of circular – permutation = nPr/2r  

Example: How many necklace of 12 beads each can be made from 18 beads of different colours?

Ans.    Here clock-wise and anti-clockwise arrangement s are same.

Hence total number of circular–permutations:       18P12/2x12

 =   18!/(6    x   24)

Restricted – Permutations

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(a)   Number of permutations of ‘n’ things, taken ‘r’ at a time,

when a particular thing is to be always included in each

arrangement

= r n-1 Pr-1

(b) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when a particular thing is fixed: = n-1 Pr-1

(c) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when a particular thing is never taken: = n-1 Pr.

(d) Number of permutations of ‘n’ things, taken ‘r’ at a time,

when ‘m’ specified things always come together = m!  x (  n-

m+1) !

(e) Number of permutations of ‘n’ things, taken all at a time,

when ‘m’ specified things always come together = n ! - [ m!

x   (n-m+1)! ]

Example:   How many words can be formed with the letters of

the word ‘OMEGA’ when:

(i)       ‘O’ and ‘A’ occupying end places.

(ii)       ‘E’ being always in the middle

(iii)       Vowels occupying odd-places

(iv)        Vowels being never together.

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Ans.

(i)    When ‘O’ and ‘A’ occupying end-places

  => M.E.G. (OA)

  Here (OA) are fixed, hence M, E, G can be arranged in  3!

ways

 But (O,A) can be arranged themselves is 2! ways.

=> Total number of words =  3!   x  2! = 12 ways.

ii)  When ‘E’ is fixed in the middle

=>    O.M.(E), G.A.

Hence four-letter O.M.G.A. can be arranged in  4!   i.e 24

ways.

(iii)   Three vowels (O,E,A,) can be arranged in the odd-

places (1st, 3rd and 5th)    =  3!    ways.

And two consonants (M,G,) can be arranged in the even-

place               (2nd, 4th) =   2 !   ways

=> Total number of ways= 3! x 2! = 12 ways.

(iv)  Total number of words   =   5!   =    120!

 If all the vowels come together, then we have: (O.E.A.), M,G

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 These can be arranged in    3!    ways.

 But (O,E.A.) can be arranged themselves in   3! ways.

 => Number of ways, when vowels come-together  =    3!  x   

3!  

= 36 ways

=> Number of ways, when vowels being never-together

= 120-36        =  84 ways.

Number of Combination of ‘n’ different things, taken ‘r’ at a time

is given by:-

nCr=  n! / r ! x (n-r)!                                

Proof: Each combination consists of ‘r’ different things, which

can be arranged among themselves in   r!  ways.

=> For one combination of ‘r’ different things, number of

arrangements =    r!

For nCr combination number of arrangements:      r    nCr

=> Total number of permutations =    r!   nCr  ---------------(1) 

But number of permutation of ‘n’ different things, taken ‘r’ at a

time

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= nPr -------(2)

From (1) and (2) :

nPr  =      r!  .  nCr

or      n!/(n-r)!  =  r!   .  nCr          

or   nCr    =       n!/r!x(n-r)!

Note: nCr  =  nCn-r

or   nCr    = n!/r!x(n-r)!   and  nCn-r  =   n!/(n-r)!x(n-(n-r))!

 =  n!/(n-r)!xr!

Restricted – Combinations

(a)  Number of combinations of ‘n’ different things taken ‘r’ at

a time, when ‘p’ particular things are always included = n-pCr-p.

(b)  Number of combination of ‘n’ different things, taken ‘r’ at

a time, when ‘p’ particular things are always to be

excluded = n-pCr

Example:      In how many ways can a cricket-eleven be chosen

out of 15 players? if

(i)  A particular player is always chosen,

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(ii)  A particular is never chosen.

Ans:   

(i)       A particular player is always chosen, it means that 10

players are selected out of the remaining 14 players.

=. Required number of ways =  14C10  = 14C4

= 14!/4!x19!  = 1365                                              

(ii) A particular players is never chosen, it means that 11

players are selected out of 14 players.

 => Required number of ways =  14C11 

 =   14!/11!x3!  = 364

(iii) Number of ways of selecting zero or more things from ‘n’

different things is given by:-   2n-1

Proof:  Number of ways of selecting one thing, out of n-things

= nC1

Number of selecting two things, out of n-things =nC2

Number of ways of selecting three things, out of n-things =nC3

Number of ways of selecting ‘n’ things out of ‘n’ things = nCn

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=>Total number of ways of selecting one or more things out of

n different things

= nC1 + nC2 + nC3 + ------------- + nCn

= (nC0 + nC1 + -----------------nCn)  - nC0

= 2n – 1                           [ nC0=1]

1. Examples:1. Suppose we want to select two out of three boys A, B, C.

Then, possible selections are AB, BC and CA.

Note: AB and BA represent the same selection.

2. All the combinations formed by a, b, c taking ab, bc, ca.3. The only combination that can be formed of three letters a,

b, c taken all at a time is abc.4. Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

5. Note that ab ba are two different permutations but they represent the same combination.

2. Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr =n!

=n(n - 1)(n - 2) ... to r factors

.(r!)(n - r!)

r!

Note:

i. nCn = 1 and nC0 = 1.ii. nCr = nC(n - r)

Examples:

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i.   11C4 =(11 x 10 x 9 x 8)

= 330.(4 x 3 x 2 x 1)

ii.   16C13 = 16C(16 - 13) = 16C3 =16 x 15 x 14

=16 x 15 x 14

= 560.3! 3 x 2 x 1

1. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? A.

564 B.645

C.735 D.7562. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together? A.

360 B.480

C.720 D.5040

3. 

In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?A.810

B.1440

C.2880

D.50400

4 Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?A.210

B.1050

C.25200

D.21400

 5. In how many ways can the letters of the word 'LEADER' be arranged?

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A.72

B.144

C.360

D.720

 

6 In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?A.159

B.194

C.205

D.209

E.None of these

7 How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated? A.5

B.10

C.15

D.20

 

8 In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women? A.266

B.5040

C.11760

D.86400

E.None of these

 

9 A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw? A.32

B.48

C.64

D.96

E.None of these

 10 In how many different ways can the letters of the word 'DETAIL'

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be arranged in such a way that the vowels occupy only the odd positions? A.32

B.48

C.36

D.60

E.120

11 In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women? A.63

B.90

C.126

D.45

E.135

12 How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?A.40

B.400

C.5040

D.2520

13 In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together? A.10080

B.4989600

C.120960

D.None of these

14 In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together? A.120

B.720

C.4320

D.2160

ENone of these

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.

Answer:1 Option D

Explanation:

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).

Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)

=

7 x 6 x 5

x6 x 5

+ (7C3 x 6C1) + (7C2)3 x 2 x 1

2 x 1

= 525 +7 x 6 x 5

x 6 +7 x 6

3 x 2 x 1 2 x 1= (525 + 210 + 21)= 756.

Answer:2 Option C

Explanation:

The word 'LEADING' has 7 different letters.

When the vowels EAI are always together, they can be supposed to form one letter.

Then, we have to arrange the letters LNDG (EAI).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (EAI) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

Answer: 3 Option D

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Explanation:

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7!

= 2520.2!

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged

in

5!= 20 ways.

3!

Required number of ways = (2520 x 20) = 50400

Answer:4 Option C

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

     = (7C3 x 4C2)

=7 x 6 x 5

x4 x 3

3 x 2 x 1 2 x 1= 210.

Number of groups, each having 3 consonants and 2 vowels = 210.

Each group contains 5 letters.

Number of ways of arranging

5 letters among themselves= 5!

= 5 x 4 x 3 x 2 x 1

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= 120.

Required number of ways = (210 x 120) = 25200.

Answer: 5Option C

Explanation:

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.

Required number of ways =6!

=

Answer:6 Option D

Explanation:

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).

Required numberof ways

= (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)

= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)= (6 x

4)6 x 5

x4 x 3

+6 x 5 x 4 x

4+

6 x 52 x 1 2 x 1 3 x 2 x 1 2 x 1

= (24 + 90 + 80 + 15)= 209.

Answer:7 Option D

Explanation:

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

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Required number of numbers = (1 x 5 x 4) = 20

Answer:8 Option C

Explanation:

Required number of ways= (8C5 x 10C6)= (8C3 x 10C4)

=

8 x 7 x 6

x10 x 9 x 8 x 7

3 x 2 x 1

4 x 3 x 2 x 1

= 11760.

Answer9: Option C

Explanation:

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

=3 x

6 x 5+

3 x 2 x 6

+ 12 x 1 2 x 1

= (45 + 18 + 1)= 64.

Answer:10 Option C

Explanation:

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.

Let us mark these positions as under:

(1) (2) (3) (4) (5) (6)

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Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.

Number of ways of arranging the vowels = 3P3 = 3! = 6.

Also, the 3 consonants can be arranged at the remaining 3 positions.

Number of ways of these arrangements = 3P3 3! = 6.

Answer:11 Option A

Explanation:

Required number of ways = (7C5 x 3C2) = (7C2 x 3C1) =

7 x 6 x

3=

63.2 x 1

Answer:12 Option C

Explanation:

'LOGARITHMS' contains 10 different letters.

Required number of words

= Number of arrangements of 10 letters, taking 4 at a time.= 10P4

= (10 x 9 x 8 x 7)= 5040.

Answer:13 Option C

Explanation:

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.

Thus, we have MTHMTCS (AEAI).

Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.

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Number of ways of arranging these letters =8!

= 10080.(2!)(2!)

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.

Number of ways of arranging these letters =4!

= 12.2!

Required number of words = (10080 x 12) = 120960

Answer:14 Option B

Explanation:

The word 'OPTICAL' contains 7 different letters.

When the vowels OIA are always together, they can be supposed to form one letter.

Then, we have to arrange the letters PTCL (OIA).

Now, 5 letters can be arranged in 5! = 120 ways.

The vowels (OIA) can be arranged among themselves in 3! = 6 ways.

Required number of ways = (120 x 6) = 720.

ASSIGNMENT Question 1 The principal wants to arrange 5 students on the platform such that the boy SALIM occupies the second position and such that the girl SITA is always adjacent to the girl RITA . How many such arrangements are possible?Question 2 When a group- photograph is taken, all the seven teachers should be in the first row and all the twenty students should be in the second row. If the two corners of the second row are reserved for the two tallest students, interchangeable only b/w them, and if the middle seat of the front row is reserved for the principal, how many such arrangements are possible?

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Question 3 If there are six periods in each working day of a school, in how many ways can one arrange 5 subjects such that each subject is allotted at least one period?Question 4 How many numbers greater than 4,00,000 can be formed by using the digits 0, 2, 2, 4, 4, 5?**Question 5 The letters of the word ZENITH are written in all possible order. How many words are possible if all these words are written out as in a dictionary? What is the rank of the word ZENITH?**Question 6 How many natural numbers not exceeding 4321 can be formed with the digits 1,2,3 and 4 if the digits can repeat?**Question 7 Three married couples are to be seated in a row having six seats in a cinema hall. If spouses are to be seated next to each other, in how many ways can they be seated? Find also the number of ways of their seating if all the ladies sit together. Question8 A boy has three library tickets and 8 books of his interest in the library. Of these 8, he does not want to borrow Chem. part II, unless Chem. Par I is also borrowed. In how many ways can be choose the three books to be borrowed?Question 9 A polygon has 44 diagonals. Find the number of its sides.** Question10 In how many ways can 10 things be equally divided b/w (i) two persons (ii) two heaps ?**Question11 In an exam. there are three multiple choice questions and each question has 4 choices. Find the no. Of ways in which a student fails to get all answer correct.Question12 In how many ways can the letters of the word PERMUTATIONS be arranged if the (i) Words start with P and end with S? (ii) Vowels are all together ? (iii) T’s are together? (iv) There is no restriction? (v) There are always 4 letters b/w P & S? Question 13 Find the no. Of perm. Of n things taken r at a time in which 2 given things always occur. In how many pers. , they are always excluded?**Question 14 There are 10 points in a plane , no three of which are in the same st. Line excepting 4 points, which are collinear. Find the (i)

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No. Of st. Lines obtained from the pairs of these points , (ii) no. Of ∆ that can be formed with the vertices as these points.Question 15 Find the no. of per. Of n different things taken r at a time such that two specific things occur together.

ANSWERS ( WITH HINTS)

Answer 1 since the boy SALIM occupies the second position, we have to arrange the remaining 4 students, according to condition two seats III, IV OR IV , V May be occupied by SITA & RITA in 4 ways, then remaining seat may be occupied by 5th student in 1 way only.So no. Of arrangements = 2.4.1=8.Answer 2 The remaining 6 teachers can be arranged in the front row in 6! Ways. (∵ middle seat is reserved for principal). The remaining 18 students can be arranged in the second row in 18! Ways. (two corner are reserved for 2 tallest , they can be occupied 2 ways) ∴ total no. Of ways = 6!.(18)!.2!Answer 3 Five subjects can be allotted 5 periods out of the six periods in 6P5 ways. Now one period is left and it can be allotted to any one of the 5 subjects in 5 ways. So, total no. Of ways = 6P5 . 5 = 3600.Answer 4 When 5 occurs the extreme left position(greater than 4) , then no. Of ways = 5 !

2! .2 ! = 30.

When 4 occurs at extreme left , then no. Of ways = 5!/2!=60 ( 0,2,2,4,5[2 appears twice]) ∴ total no. Of ways 90.Answer 5 (i) total no. Of possible words = 6!=720 (ii) The no. Of words beginning with E = 5! = 120 , similarly for H ,I, N or T is 120. The words start with Z will also be 120 and will have their rank from 601 to 720.Of these 120 words start with Z, the no. Of words with E in the second place is 120/5 = 24. ZE**** will have their rank from 601 to 624. Of these 24 words , NO. Of words with H(or I or N or T) in the third place is 24/2=6

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Rank order : ZEH*** , ZEI***, ZEN*** 601 to 606, 607 to 612, 613 to 618 resp., start with ZEN are 613...ZENHIT, 614... ZENHTI, 615... ZENIHT, 616... ZENITH.Answer 6 1- digit nos. = 4, 2-digit nos. = 4x16, 3-digit nos. = 64, 4-digit nos. 256. Out of 256 , let us discard those which are greater than 4321. (i) 4 at thousand’s place & hundred’s place= 1.1.4.4=16 ,

4 4 1234

1234

Similarly (ii) 4 at thousand’s place ,3 at hundred’s place, 3 or 4 in ten’s = 1.1.2.4=8, (iii) 4 at thousand’s place ,3 at hundred’s place, 2 in ten’s & 2,3,4 at unit’s= 1.1.1.3=3. Total 4-digit no.greater than 4321 are = 27, answer is 256-27=313.Answer 7(i) Let A, B, C be three married couples can be seated = 3! & can sit in 2 ways =2! ∴ req. Ways of seating = 3!.2!.2!.2!=48 . (ii) three ladies can be seated in 4 ways which are at seat number (1,2,3), (2,3,4), (3,4,5), (4,5,6). They can interchange their seats =3! Ways. Men can be seated at three remaining seats in 3! , so req. No. Of ways = 4.3!.3!= 144 .Answer 8 case-1 he borrows chem.. part II, chem.. part I and one more book out of the remaining 6 (8-2) books. Case-2 he does not borrow chem.. part-II and , so borrow all 3 books out of 7 ∴ total no. Of ways = 6C1 + 7C3=41. Answer 9 No. Of diagonals = nC2 - n = 44 ( nC2 = no.of st. Lines of polygon n sides).Answer 10 (i) 10 things be equally divided b/w two persons, groups are distinct , 10!/ (5!)2 = 252 (ii) 10 things be equally divided b/w two heaps no distinction can be made = ½ (252)=126. Answer 11 Req. No. Of ways= 4.4.4 – 1=63. Answer 13 (i) 10!/2! (ii) consider 5 vowels as one letter (8!/2!) .(5!) (iii) consider 2 T’s as one letter = (11)! (iv) 12!/2!

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(v) 12 letters in 12 places as P****S****** leaving 4 places in between , Thus P & S may be filled up in 7 ways, same for S & P , remaining 10 letters are arranged in 10!/2! Ways ∴ req. No. of ways 10!/(3!.2!.2!) = 151200. Answer 14 No. Of per. Of n things taken r at a time is same as the no. Of ways of filling up r places with n things, we arrange 2 things which can be arranged in r places in rP2 ways. Also remaining (r-2) places can be filled up with remaining (n-2) things in n-2Pr-2 ways.No. Of per. (2 are included) = rp2 . n-2Pr-2 , No. Of per. (2 are not included) = n-2Pr

Answer 15 (i) No. Of st. Lines formed by joining 10 points , taking 2 at a time = 10C2 = 45. No. Of st. Lines joining 4 points = 4C2 = 6 , but 4 are collinear , when join pairwise give only one line. Req. No. of lines = 45 – 6 + 1=40 (ii) No. of ∆s formed by joining the points , taken 3 at a time = 10C3 = 120, no. of ∆s formed by joining 4 points , taken 3 at a time = 4C3 = 4 Req. No. of ∆s = 120 – 4 = 116.Answer 16 Two specific things can be in r places in (r-1) ways and two things can be arranged among themselves in 2! Ways and the remaining (n-2) things will be arranged in (r-2) places in n-2Pr-2 ways. Req. Per. = 2! (r-1) . n-2Pr-2 .