Class x English Maths Chapter04

This unit facilitates you in,

stating the fundamental principle of

counting.

defining permutation and combination.

using factorial notation.

deriving the formula for number of

permutations (nPr).

deriving the formula for number of

combinations (nCr).

differentiating between permutations and

combinations

solving problems using nPr and nC

r.

Fundamental principle of

counting

Permutation

Factorial notation

!

!

n

r

nP

n r

Combination

!

! !

n

r

nC

r n r

4

Jacob Bernoulli

(1654 - 1705 A. D.)

Swiss mathematician, Jacob

Bernoulli has given a complete

treatment of permutations

and combinations in his book

‘Ars Conjectandli’ (Posthumously

published in 1713 A. D)

In mathematics the art of proposing a question

must be held of higher value than solving it.

- George Cantor

Permutations and

Combinations

66 UNIT-4

Let us begin this unit with some real life situations. Study them and discuss in groups.

Illustration 1

We see different types of motor vechicles plying on the road. Each and every vehicle will

have a number plate with it's registration number.

Recall that each number plate will have the following information.

1) Name of the state (Eg : KA - for Karnataka)

2) Regional transport office code number.

(05 - Bangalore - Jayanagar RTO)

3) English alphabet/s

4) Followed by a number (from a single digit to a four digit number)

Do you know how many vehicles can be registered in any one of the series? You

know that there can be 9999 vehicles!!

Have you ever thought how these numbers are generated by the R.T.O? What

mathematical principle is behind this calculation?

Illustration 2

Suppose we have a suitcase with number lock. The number lock has three wheels each

labelled with ten digits from 0 to 9. The lock can be opened if three specific digits are

arranged in a particular arrangement or may be in a sequence say with no repetitions.

Let us suppose that we have forgotten this sequence of digits. In order to open the lock,

how many sequences of three - digits we have to check? To answer this question we have

to check all the three - digit numbers. But this will take lot of time. Is there any

mathematical idea which can help us to known the number of possibilities?

In this unit let us learn some basic counting techniques by which we are able to

find solutions for the above situations.

As a first step, we shall examine carefully a principle which is most fundamental to

learning of the counting techniques.The main subject of this unit is counting. Given a

set of objects, the task is to arrange a subset according to some specifications or to select

a subset as per some specifications.

Fundamental Principle of Counting (FPC)

Study the following illustrations.

Example 1: There are three dolls as shown.

Two out of them have to be arranged on a shelf. In how many different ways can they be

arranged? How do we count this?

Observe the tree diagram representing the number of ways of selecting the first

doll and the second doll.

KA05

EV4007

Permutations and Combinations 67

From this diagram, we can conclude that the first doll can be

selected in three different ways and for each of these first selection,

the second doll can be selected in two different ways.

So there are totally 6 ways.

The total number of ways of arranging two dolls out of three

is 3 × 2 = 6 ways.

The 6 different ways can be represented as follows:

Example 2: A boy has 2 pants and 4 shirts. How many different pairs of a pant and a shirt

can he dress up with?

There are two ways in which a pant can be chosen and for every choice of a pant

there are four ways in which a shirt can be chosen.

Let us denote the two pants as P1 and P

2 and the shirts as S

1, S

2, S

3 and S

4. The

different ways of pairing a pant and a shirt can be represented by a tree diagram as shown

below.

there are totally 8 ways.

In other words, there are 2 × 4 = 8 ways of pairing a pant and a shirt.

Example 3: Sanjay goes to a hotel to have breakfast.

The menu card indicates the following items that are served:

Tiffin Sweets Hot drinks

Idli Kesari bath Coffee

Dosa Jamoon Tea

Poori Badam milk

Kharabath

In how many different ways can he select one item from each type?

Tiffin can be chosen in 4 different ways. After a tiffin is chosen, a sweet can be

chosen in 2 different ways.

there are 4 × 2 = 8 ways in which a tiffin and a sweet can be chosen.

Pants

P1

P2

S ––– PS

1 1

1

S ––– PS2 1

2

S ––– PS3 1

3

S ––– PS4 1

4

S ––– PS1 2

1

S ––– PS2 2

2

S ––– PS3 2

3

S ––– PS4 2

4

Shirts

68 UNIT-4

For each of the above 8 ways, a hot drink can be chosen in 3 different ways.

there are 8 × 3 = 24 ways in which Sanjay can choose the three items.

the total number of ways = 4 × 2 × 3 = 24

If we represent tiffin as T1, T

2, T

3 and T

4, sweets as S

1 and S

2, hot drinks as H

1, H

2

and H3, the 24 possible ways can be listed in the table as follows.

T1 S

1 H

1T

2 S

1 H

1T

3 S

1 H

1T

4 S

1 H

1

T1 S

2 H

1T

2 S

2 H

1T

3 S

2 H

1T

4 S

2 H

1

T1 S

1 H

2T

2 S

1 H

2T

3 S

1 H

2T

4 S

1 H

2

T1 S

2 H

2T

2 S

2 H

2T

3 S

2 H

2T

4 S

2 H

2

T1 S

1 H

3T

2 S

1 H

3T

3 S

1 H

3T

4 S

1 H

3

T1 S

2 H

3T

2 S

2 H

3T

3 S

2 H

3T

4 S

2 H

3

The data from the above three examples is entered in the table. Study the table and

discuss in groups to infer the principle of counting.

Number of ways Total number

Activity 1 Activity 2 Activity 3 of ways

1 3 2 – 3 × 2 = 6

2 4 2 – 4 × 2 = 8

3 4 2 3 4 × 2 × 3 = 24

We can conclude and state the fundamental principle of counting (FPC) or

multiplication principle as follows :

If one activity can be done in 'm' number of different ways and corresponding to each of

these ways of the first activity, second activity (independent of first activity) can be

done in 'n' number of different ways then, both the activities, one after the other can be

done in (m × n) number of ways.

The above principle can be generalised for any finite number of events.

For three activities, the FPC can be stated as:

If one activity, can be done in 'm' number of different ways, for each of these 'm'

different ways, a second activity can be done in 'n' number of different ways and for

each of these activities, a third activity can be done in 'p' ways, then all the three

activities one after the other can be done in (m × n × p) number of ways.

We can now easily find the total number of ways of doing more than two activities

without actually listing, counting or drawing tree diagram. We shall illustrate this FPC

with a few more examples.

ILLUSTRATIVE EXAMPLES

Example 1: How many 2 - digit numbers can be formed from the digits {1, 2, 3, 4, 5}

without repetition and with repetition?

Sol: The digits in the selection set are {1, 2, 3, 4, 5}.

Since we have to form 2 - digit number, let us draw 2 boxes, one for

units place and another for tens place.

Example

T U


We fill the units place first and then tens place.

The box for units place can be filled in 5 different ways with the digits 1, 2, 3, 4

and 5.

After filling the units place we are left with 4 digits, as repetition of digits are not

allowed, the box for tens place can be filled in 4 different ways.

By applying fundamental principle of counting we get 5 × 4 = 20 ways.

there are 20 ways of forming 2 - digit numbers by using 1, 2, 3, 4, 5 without

repetition.

If the repetition of digits are allowed then the tens place can also be filled by 5

different ways, using 1, 2, 3, 4 and 5.

The total number ways a 2 - digit number can be formed by using 1, 2, 3, 4 and 5

with repetition is 5 × 5 = 25 ways.

Note: Discuss in groups and verify the answer by actually listing the 2 - digit numbers.

Example 2. How many 3 letter code can be formed by using the five vowels without

repetitions?

Sol. The vowels are a, e, i, o and u.

We have to form a three letter code.

The first letter can be chosen in 5 ways.

The second letter can be chosen in 4 ways. ( no repetition is allowed)

The third letter can be chosen in 3 ways. ( no repetition is allowed)

The total number of ways in which the 3 letter code can be formed without repetition

= 5 × 4 × 3 = 60 ways

Example 3. How many 3 - digit numbers can be formed from the digits 0, 1, 2, 3 and 4 with

repetitions?

Sol. A Three - digit number will have three places - Hundreds, tens, units.

The digits in the selection set are {0, 1, 2, 3, 4}

Hundreds place - Zero cannot occupy hundreds place, because it becomes a 2 - digit

number. Hence, the hundreds place can be filled in by 4 ways, i.e. 1, 2, 3, 4.

Tens place - We have five ways of filling tens place, as repetitions are allowed and

zero can also be used.

Units place - This can also be filled in 5 different ways.

the total number of 3 - digit numbers that can be formed using

0, 1, 2, 3, 4 are 4 × 5 × 5 = 100 numbers.

Example 4. Now let us try to solve the suitcase lock

problem (illustration 2 in page 2), by applying the

fundamental principle of counting.

Sol. The number lock of the suitcase has three digits.

Each number can be formed in 10 different ways

using the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.

6 2 07 3 18 4 2

70 UNIT-4

It is also evident that the digits can be repeated.

the total number of ways the number lock code can be formed is 10 × 10 × 10 =

1000 ways

So, to open the suitcase 1000 different ways have to be checked.

EXERCISE 4.1

1. How many 3 - digit numbers can be formed using the digits 1, 2, 3, 4, 5, 6 without

repeating any digit.

2. How many 3 digit even numbers can be formed using the digits 3, 5, 7, 8, 9, if the

digits are not repeated?

3. How many 3 letter code can be formed using the first 10 letters of English alphabet,

if no letter can be repeated?

4. How many 5 digit telephone numbers can be formed using the digits 0 to 9, if each

number starts with 65 and no digit appear more than once?

5. If a coin is tossed 3 times, find the number of outcomes.

6. Given 5 flags of different colours, how many different signals can be generated if

each signal requires the use of 2 flags one below the other?

Permutations and Combinations

While learning the fundamental principle of counting, we have observed that in all the

examples some number of objects are given and out of these a few objects are to be chosen.

Now, let us study more about the selection and arrangement of things.

Study the examples given in the table.

Read the first problem in column 'P' and then the first problem in column 'C' Now compare

them. Repeat the same procedure for the remaining set of problems.

Do you notice any similarties in the problems of each column? For the problems in the

column 'P', we have to find out the number of ways of arrangements. For the problems in

the column 'C' we have to find the number of ways of selections.

Column 'P'

3 persons contest for the post of a

president and a secretary. In how

many ways they can be elected?

A child has 8 frocks and 4 pairs of

shoes. In how many ways the child

can dress herself?

In how many ways can all the

letters of the word TEACH be

arranged?

Column 'C'

There are 3 condidates contesting for

2 posts. In how many ways can the

posts be filled.

There are 8 white and 4 red roses in

a garden. In how many ways 4 flowers

of which 2 are red be plucked?

In how many ways all the letters of

the word MEANS can be selected?

Sl No.

1.

2.

3.


In solving prolems of the above types the first step is we must carefully identify,

whether it is the arrangements or selections. As an example let us understand the first

problem in the above table.

It is also evident from the above discussion that, in examples of column P the

arrangement of things or objects or persons is done with regard to order. In examples of

column C, the selection of things or objects or persons is done without regard to order.

Such arrangements with regard to order are called permutations and selections

without regard to order are called combinations.

Permutation : A permutation is an ordered arrangement of a set of objects.

It is an act of arrangements of objects in an orderly way.

A permutation of n elements taken r at a time is any ordered subset of r elements

taken from the set of n elements.

The number of permutations of 'n' different objects taken 'r' at a time is denoted asnP

r where r n.

Combination : A combination is a selection of a set of objects without any order. It is an

act of selection of objects not involving any orderly way.

A combination of n elements taken r at a time is any subset of r elements taken from

the set of n elements (without regard to order).

Note that in a set there is no order for listing the elements. e.g. {1, 3, 2} = {2, 1, 3}

The number of combinations of 'n' different objects taken 'r' at a time is denoted asnC

r, where r n.

Let the 3 persons be A, B and C. There

are two posts, president and a secretary.

Let us have the box notation and write all

the possible ways.

Case 1. If 'A' is the president then 'B' can

be secretary or 'C' can be secretary

Case 2 : If 'B' is the president then 'A' can

be secretary or 'C' can be secretary

Case 3: If 'C' is the president then 'A' can

be secretary or B can be the secretary.

So all together there 6 possible ways of

electing.

Let the 3 persons be A, B and C. There are

two posts only. Let us have the box notation

and write all the possibilities.

Choices for posts

Case 1. If 'A' is selected for one post,

another post can be given to B or C.

Case 2 : If 'B' is selected for one post then

the other post can be given to C but not to

A because BA and AB are the same

selections. Hence there can be 3 ways by

which the posts can be filled up

SLNo. Post - 1 Post - 2

1

2

3

A

A

B

B

C

C

President Secretary

1

2

3

4

5

6

A

A

B

B

C

C

B

C

A

C

A

B

72 UNIT-4

EXERCISE 4.2

I. Below are given situations for arrangements and selections. Classify them as

examples of permutations and combinations.

1. A committee of 5 members to be chosen from a group of 12 people.

2. Five different subject books to be arranged on a shelf.

3. There are 8 chairs and 8 people to occupy them.

4. In a committee of 7 persons, a chairperson, a secretary and a treasurer are to be

chosen.

5. The owner of children's clothing shop has 10 designs of frocks and 3 of them have to

be displayed in the front window.

6. Three-letter words to be formed from the letters in the word 'ARITHMETIC'.

7. In a question paper having 12 questions, students must answer the first 2 questions

but may select any 8 of the remaining ones.

8. A jar contains 5 black and 7 white balls. 3 balls to be drawn in such a way that 2 are

black and 1 is white.

9. Three-digit numbers are to be formed from the digits 1, 3, 5, 7, 9 where repetitions

are not allowed.

10. Five keys are to be arranged in a circular key ring.

11. There are 7 points in a plane and no 3 of the points are collinear. Triangles are to

be drawn by joining three non-collinear points.

12. A collection of 10 toys are to be divided equally between two children.

In each of the above examples give reason to explain why it is permutation or

combination.

Try: List atleast 5 examples each for permutation and combination. Discuss in the

class.

To find the general formula for the number of ways by which rth box can be filled up:

By using FPC we can write the expansion of each permutation as follows :-

In each permutation, observe

carefully the number of ways in

which the last box is filled up.

There is some common pattern in all

the permutations.

There is a relationship between

values of n, r and the number of

ways in which the last box can be filled up.

The relation is as follows :

55P 5 4 3 2 1

54P 5 4 3 2

53P 5 4 3

52P 5 4

51P 5


'Subract 'r' from 'n' and add one to it' to get the number of ways by which rth box can be filled

up i.e., (n – r + 1)

Note that when use say rth box, r is atleast 1. So r > 0

Study the following examples:

1) Consider 5P5 - we know that the last box i.e., 5th box can be filled up in just 1 way. We

can get this by subtracting (5 from 5) and add 1 to it.

i.e., (5 – 5 + 1) = 0 + 1 = 1 way.

2) Consider 5P4 - Here also we know that the last box i.e., 4th box can be filled up in 2

ways. We can get this by subtracting (4 from 5) and add 1 to it.

i.e., (5 – 4 + 1) = 1 + 1 = 2 ways.

3) Consider 5P3 - Here also we see from the above table last box can be filled up in 3

ways. This we can get by subtracting (3 from 5) and then add 1 to it.

i.e., (5 – 3 + 1) = 2 + 1 = 3 ways

Hence, we can conclude that in nPr the rth box can be filled in (n – r + 1) ways.

To find the number of permutations of 'n' distinct objects taken 'r' at a time where

0 r n and the objects do not repeat.

The number of permutations of 'n' objects taken 'r' at a time is same as the no of ways of

filling 'r' blank boxes with 'n' given objects.

Consider 'n' different objects and 'r' blank boxes as shown below.

1 2 3 r –1 r

nways

( 1)

ways

n ( 2)

ways

n [ – 2)]n r [ –n r 1)]ways ways

The first box can be filled by 'n' number of ways by putting 'n' different given objects.

Thus, there are 'n' different ways of filling up the first box.

After filling up the first box by any one of the 'n' objects, we are left with (n – 1)

objects as repetitions are not allowed.

So, the second box can be filled up in (n – 1) number of ways.

Thus, the first 2 boxes can be filled up in n (n – 1) number of ways (by FPC).

Now after filling up the first 2 boxes, we are left with (n – 2) objects.

Thus, the 3rd box can be filled up in (n – 2) number of ways.

Again by F.P.C the first three boxes can be filled up in n(n – 1) (n – 2) ways. Observe

that a new factor is introduced with each new box filled up, and that at any stage the

number of factors is the same as the number of boxes filled up. Hence, by FPC the number

of ways of filling 'r' boxes in succession is given as follows.

n (n – 1) (n – 2) ............................. [n – (r – 1)]

74 UNIT-4

Thus the number of permutations of 'n' objects taken 'r' at a time is given by nPr

nPr = n (n – 1) (n – 2) ............................. [n – (r – 1)]

Corollary : The number of permutations of 'n' objects taken all at a time is

nPn = n (n – 1) (n – 2) ............................. [n – (n – 1)]

nPn = n (n – 1) (n – 2) ............................. 1

or nPn = n (n – 1) (n – 2) ............................. 3 2 1 ('n' factors)

Factorial notation :

Observe that the RHS of the above expresssion is the product of first n natural numbers.

It is denoted by a notation n!. n! is read as n factorial.

n (n – 1) (n – 2) ............................. 3 2 1 = n!

n! denotes the product of first n natural numbers.

Thus 4! = 1 × 2 × 3 × 4 or 4 × 3 × 2 × 1

3! = 1 × 2 × 3 or 3 × 2 × 1

2! = 1 × 2 or 2 × 1

n! = 1 × 2 × 3 ×..... × n or n × (n – 1) (n – 2) × .... × 3 × 2 × 1

0! by definition is taken as 1.

[Note : If n is a negative number or a decimal, n! is not defined.]

Remember :

n! is the product of first 'n' natural numbers. n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1

Sn is the sum of first 'n' natural numbers. S

n = 1 + 2 + 3 + ........... + n

Study the expansions of factorial notations given below.

n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1

(n – r)! = (n – r) (n – r – 1) (n – r – 2) × ...... × 3 × 2 × 1

(n – r + 1)! = (n – r + 1) (n – r) (n – r – 1) × ...... × 3 × 2 × 1

(n – r – 1)! = (n – r – 1) (n – r – 2) (n – r – 3) × ...... × 3 × 2 × 1

(r – 1)! = (r – 1) (r – 2) (r – 3) × ...... × 3 × 2 × 1

In general ; n! = n (n – 1) (n – 2) × ...... × 3 × 2 × 1

n! = n [(n – 1) (n – 2) × ...... × 3 × 2 × 1]

n! = n [(n – 1)!]

or n! = n (n – 1) (n – 2)!

or n! = n (n – 1) (n – 2) (n – 3)! and so on.

For example, Consider 5!

5! = 5 × 4 × 3 × 2 × 1

5! = 5 × (4!)

5! = 5 × 4 × (3 !)

5! = 5 × 4 × 3 × (2!)


EXERCISE 4.3

1. Convert the following products into factorials.

(i)1 × 2 × 3 × 4 × 5 × 6 × 7 (ii) 18 × 17 × ........ × 3 × 2 × 1

(iii) 6 × 7 × 8 × 9 (iv) 2 × 4 × 6 × 8

2. Evaluate : (i) 6! (ii) 9! (iii) 8! – 5!

(iv)7!

5!(v)

12!

9! 3!(vi)

30!

28!

3. Evaluate : (i)!

!

n

n r (ii)

!

! !

n

n r r when n = 15 and r = 2

4. Find the LCM of 4!, 5!, 6!

5. (i) If (n +1)! = 12 (n – 1)! find the value of n

(ii) If 1 1

9! 10! 11!

n find the value of n

6. Simplify : ( 1)!

( 2)! (n–1)

n

n

To derive the formula for nPr in factorial notation

By Fundamental Principle of counting, nPr = n (n – 1) (n – 2) × .......... × (n – r + 1)

Consider the RHS of the equation,

n (n – 1) (n – 2) ........ (n – r + 1)

The first factor of the RHS is n. The last factor of the RHS is (n – r + 1).

RHS is the product of natural numbers starting from 'n' in descending order upto

(n – r + 1). Had the product continued upto 1 what we would have got? We would have got n!

What extra factor we have to write to get n! ?

[(n – r) (n – r – 1) × ...... × 3 × 2 × 1] which is (n – r)!

nPr = n (n – 1) (n – 2) × ..... × (n – r + 1)

nPr =

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 ! nP

r =

n!

n r ! [If r > 0]

Number of permutations of 'n' things taken 'r' at a time is !

!

n

r

nP

n r, where 0 r n.

The above formula for nPr holds good only when repetitions are not allowed.

76 UNIT-4

The number of permutations of 'n' different objects taken 'r' at a time, where repetition

is allowed is nr.

Now let us study some special cases of nPr

Case 1: What happens when r = 0? We shall see it now.

If r = 0 in how many ways can we arrange 0 objects from n objects?

Arranging no object at all is the same as leaving behind all the objects as they are

and we know that there is only one way of doing it. Thus 0 1n

P

n

oP = n!

n 0 !=

n!

n!= 1

Example : 1, 1, 1100 250 958

0 0 0P P P

Case 2: Let r = n

nPr = n (n – 1) (n – 2 ) × .................... × (n – r + 1)

nPn = n (n – 1) (n – 2 ) × .................... × (n – n + 1)

nPn = n (n – 1) (n – 2 ) × .................... × 1

nPn = n (n – 1) (n – 2 ) × .................... × 3 × 2 × 1

nPn = n!

Also,! ! !

!! 0! 1

n

n

n n nn

n nP (recall the definition 0! = 1)

Example : 1009P1009

= 1009 !, 10P10

= 10 !, 1400P1400

= 1400 !

Case 3: Let r = 1

!

!

n

r

n

n rP 1

!

1 !

n nP

n1

1 !n

n nP

1 !n nP

1 = n

Example :100P1 = 100, 457P

1 = 457

Discuss: 5P6 is meaningless. Why?

Remember: nPr

!

!

n

n r

If r = 0 then nP0 = 1 If r = n then nP

n = n! If r = 1 then nP

1 = n



Example 1. Evaluate : (i) 7P3

(ii) 8P5

Sol. (i) 7

3

7! 7! 7 6 5 4!

7 3 ! 4!P

4!7 6 5 210

(ii) 8

5

8! 8! 8 7 6 5 4 3!

8 5 ! 3!P

3!8 7 6 5 4 6720

Example 2. Find 'r ', if 5 4Pr = 6 5P

r – 1

Sol.4! 5!

5 64 ! 5 1 !r r

5! 6 5!

4 !r 6 !r

1

4 !r

6

6 5 4 !r r r

(6 – r) (5 – r) = 6 r2 – 11r +24 = 0 (r – 8) (r – 3) = 0

r = 3 [r = 8 has no meaning]

Example 3. Prove that n! + (n + 1)! = n! (n + 2)

Sol. LHS: n! + (n + 1)! = n! + (n + 1) n! = n! (1 + n + 1)

= n! (n + 2) = RHS

Example 4. Find 'n' if 4

1

4

5

3

n

n

P

P

Sol.

( 1)n n 2n 3n

( 1)n 2n 3n

5 5,

3 4 34

n

nn

5(n – 4) = 3n 5n – 20 = 3n 2n = 20

n = 10

Example 5. If 2n + 1Pn – 1

: 2n – 1Pn = 3 : 5 find n.

Sol.

2 1

1

2 1

3

5

n

n

n

n

P

P

i.e., 5 × 2n + 1Pn – 1

= 3 × 2n – 1Pn

5. 2 1 ! 3. 2 1 !

2 1 !2 1 1 !

n n

n nn n

5 2 1 2 2 1n n n ! 3 2 1

2 !

n

n

!

1 !n

10 2 1

2 1 1

n n

n n n n

3

! 1n !

78 UNIT-4

10 (2n + 1) = 3(n + 2) (n + 1)

3n2 – 11n – 4 = 0 (3n + 1) (n – 4) = 0

1

3n or n = 4.

Since n is a positive integer, n = 4.

Example 6. If nPn = 5040, find 'n'

Sol. nPn = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

n! = 7! n = 7

(ii) If nP2 = 90 find 'n'

Sol. nP2 = 90

n(n – 1) = 10 × 9

n = 10

(iii) If 11Pr = 990 find 'r'

Sol. 11Pr = 990

11Pr = 11 × 10 × 9

r = 3

EXERCISE 4.4

1. Evaluate:

i) 12P4

ii) 75P2

iii) 8P8

iv) 15P1

v) 38P0

2. 1. If nP4 = 20 nP

2 find 'n'

2. If 16 nP3 = 13 n+1P

3 find 'n'

3. If 5Pr = 2 .6P

r –1 find 'r'

3. 1. If nP4 : nP

5 = 1 : 2 find 'n'

2. If n–1P3 : n+1P

3 = 5 : 12 find 'n'

4. 1. If 9P5 + 5 9P

4 = 10P

r, find 'r'

2. If 10Pr+1

: 11Pr = 30 : 11, find 'r'

5. 1. Prove that n+1Pr+1

= (n + 1) nPr

2. Show that 10P3 = 9P

3 + 3 9P

2

50407

7206

1205

244

63

22

1

11 990

10 90

9 9

1


Practical Problems on Permutation

Example 1. 6 songs are to be rendered in a programme. In how many different orders

could they be performed?

Sol. The different orders in which 6 songs that can be performed is 6P6 = 6!

6! = 6 × 5 × 4 × 3 × 2 × 1 [FPC]

= 720 different orders

or!

!

n

r

nP

n r

6

6

6!

6 6 !P =

6 5 4 3 2 1

1 = 720

6 songs can be rendered in a programme in 720 different orders.

Example2. How many words (with or without dictionary meaning) can be made from

the letters in the word LASER assuming that no letter is repeated it, such that

(i) All letters are used at a time

(ii) 3 letters are used at a time

(iii) All letters are used such that it should begin with letter A and end with letter R

Sol. Number of letters in the word LASER is 5.

(i) If all letters are used at a time, then n = 5, r = 5

The number of words that can be formed

=5

5

5! 5! 5!5! 120

5 5 ! 0! 1P

120 words can be formed with or without meaning.

(ii) If 3 letters are used at a time, then n = 5, r = 3

The number of words formed with or without meaning

=5

3

5! 5 4 3 2

5 3 !P

1

2 160

So there will be 60 words with or without meaning.

(iii)

There are 5 letters in the given word LASER of which two letters A and R are

fixed in the first & last place respectively.

So only three boxes have to be filled namely 2nd, 3rd and 4th. This can be done in

3

3

3! 3 2 1 66ways

3 3 ! 0! 1P

There will be only 6 words with or without meaning which begins with letter

'A' and end with the letter 'R'.

Box 1 Box 2 Box 3 Box 4 Box 5

letter A is fixed letter R is fixed

80 UNIT-4

A = Adenine

C = Cytosine

G = Genanine

T = Thymine

Example 3. In how many ways can 7 different books be arranged on a shelf? In how

many ways three particular books are always together?

Sol. The 7 books can be arranged in

7P7 = 7!

= 7 × 6 × 4 × 3 × 2 × 1

= 5040 ways.

Since three particular books are always together, let us tie three books together

and then consider them as one book (or one unit). Remaining four books have to be

considered separately. So in all we can consider 7 books as

[4 books + 1 book ] = 5 books

(1 unit of 3 books)

These 5 books can be arranged in 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways.

In each of these 120 ways, the three particular books can be arranged in

3! = 3 × 2 × 1 = 6 ways { (B1 B

2 B

3)(B

1 B

3 B

2), (B

2 B

1 B

3), (B

2 B

3 B

1), (B

3 B

1 B

2)(B

3 B

2 B

1)

By FPC, the total number of arrangements 5P5 × 3! = 120 × 6 = 720 ways.

So, the number of ways of arranging 7 books, so that three particular books is never

together is 5040 – 720 = 4320

Example4. How many five digit numbers can be formed with the digits {2, 3, 5, 7, 9}

which lie between 30,000 and 90,000 using each digit only once.

Sol. Ten thousand Thousand Hundred Tenth unit 's

place place place place place

3 {3,5 7} 4 3 2 1ways or ways ways ways way

Since we require numbers which are greater than 30,000 and less than 90,000,

the ten thousandth place can be filled by {3, 5 or 7} only.

Therefore ten thousandth place can be filled in by 3 ways only. The remaining four

places can be filled with the remaining four digits. This can be done is 4! ways.

Therefore the total number of numbers lying between

30,000 and 90,000 is 3 × 4! { FPC}

= 3 × 4 × 3 × 2 × 1 = 72

Example 5. A DNA molecule will have a nitrogen base which

consists of different bases A, G, T or C all attached to it? In

how many ways the three bases can be arranged without

repetition.

Sol. Here, n = 4 and r = 3. Hence the number of ways

4

3

4! 4 3 2 124 ways

4 3 ! 1P


EXERCISE 4.5

1. How many words with or without dictionary meaning can be formed using all the

letters of the word 'JOULE' using each letter exactly once?

2. It is required to seat 5 men and 4 women in a row so that the women occupy the

even places. How many such arrangements are possible?

3. In how many ways can 6 women draw water from 6 wells, if no well remains unused?

4. Seven students are contesting the election for the Presidentship of the student's

union. In how many ways can their names be listed on the ballot papers?

5. 8 students are participating in a competition. In how many ways can the first three

prizes be won?

6. Find the total number of 2- digit numbers.

7. There are 5 stickers of different sizes. It is desired to make a design by arranging

them in a row. How many such designs are possible?

8. How many 4- digit numbers can be formed using the digits 1, 2, 3, 7, 8, 9 (repetitions

not allowed)?

(a) How many of these are less than 6000?

(b) How many of these are even? (c) How many of these end with 7?

9. There are 15 buses running between two towns. In how many ways can a man go to

one town and return by a different bus?

To find out the number of combinations of 'n' distinct objects taken 'r' at a time where

0 r nConsider the following examples of selecting and arranging letters from a, b, c and d.

Here n = 4:

Combinations Permutations

r = 1,

a, b, c, d a, b, c, d4C

1 = 4 4P

1 = 4

Each combination of 4C1 give rise to 1! Permutation.

r = 2 ab,ac,ad,ba,bc,bd

ab,ac,ad,bc,bd,cd ca,cb,cd,da,db,dc4C

2 = 6 4P

2 = 12

Each combination of 4C2 give rise to 2! Permutations.

r = 3 abc,abd,acd,bcd abc abd bcd acd

4C3 = 6 acb adb bdc adc

bac bad cbd cad

bca bda cdb cda

cab dab dbc dac

cba dba dcb dca

4P3 = 24 ways

Each combination of 4C3 give rise to 3! Permutations.

82 UNIT-4

r = 4 abcd abcd bacd cabd dabc4C

4 = 1 abdc badc cadb dacb

acbd bcad cdba dbac

acdb bcda cdab dbca

adcb bdac cdab dcab

abdc bdca cbda dcba

4P4: 24

Each combination in 4C4 gives rise to 4! Permutations.

In General:

Each combination in nCr give rise to r! permutations.

How many permutations can there be altogether?

Total no of Permutations = r ! × nCr

nPr = r ! × nC

r nC

r =

!

n

rP

r

i.e. nCr =

!

! !

n

n r r

Therefore, the number of combinations of 'n' distinct objects, taken 'r' at a time is

given by

n

r

n!C

r! n - r !

Note: The relationship between nPr and nC

r is given by nC

r =

!

n

rP

r

Study the various expansions of nCr

1. r

!

nn

r

PC

r

2. !

! !

n

r

nC

r n r

3.

1 2 ............... 3 2 1

! !

n

r

n n nC

r n r

4.1 2 ................ 1

!

n

r

n n n n rC

r

Now let us consider some special cases of nCr

Case (i): Let r = 0

If r = 0 then, 0

! !1

0! 0 1 !

n n nC

n n nC

0 = 1

100C0 = 1, 500C

0 = 1, 1000C

0 = 1


Case (ii): Let r = 1

If r = 1 then, 1

!

1! 1 !

n nC

n=

1n n !

1 1n !

nC1 = n

Example: 100C1 = 100, 200C

1 = 200 357C

1 = 357

Case (iii): Let r = n

If r = n then, ! !

! ! !0!

n

n

n nC

n n n n

!

1! 1

n

n

nCn = 1

Example: 100C100

= 1, 789C789

= 1, 1497C1497

= 1

Remember :

!

! !

n

r

nC

n r r

If r = 0, nC0 = 1, If r = 1, nC

1 = n, If r = n, nC

n = 1


Example 1. Evaluate : (i) 6C4

(ii) 8C3

Sol.!

! !

n

r

nC

n r r

(i)6

4

6! 6!

6 4 !4! 2!4!C

3 6 5 4!

2 1 4!15

(ii)8

3

8! 8!

8 3 !3! 5!3!C

4 8 27 6 5!

5! 31 256

1

Example2.If nC2 = 10, find n

Sol. nC2 = 10

!

! !

n

r

nC

n r r 2

!

2 !2!

n nC

n

1 2 !n n n

2 !n 2!

2

110

2

n n nC

n(n – 1) = 20 5(5 – 1) = 20 5 × 4 = 20 n = 5

84 UNIT-4

Example 3. If 6Pr = 360 and 6C

r = 15, find 'r'.

Sol. 6Pr = 6C

r × r !

360 = 15 × r !

360

! 2415

r = 1 × 2 × 3 × 4 = 4!

r = 4

Example 4. Prove that nCr = nC

n – r

Sol. We know that !

C! !

n

r

n

r n r.........(i)

Replacing r by (n – r) in the equation (i) we get

!C

! !

n

n r

n

n r n n r

!C

!

n

n r

n

n r n n !r

!C

! !

n

n r

n

n r r ........(ii)

By comparing (i) and (ii) we can say that

nCr = nC

n – r

Example 5. If nC8 = nC

12, find n.

Sol. nC8 = nC

12

nC8 = nC

n –12n n

r n rC C

8 = n –12 n = 12 + 8 = 20.

EXERCISE 4.6

1. Evaluate (i) 10C3

(ii) 60C60

(iii) 100C97

2. (i) If nC4 = nC

7 find n. (ii) If nP

r = 840, nC

r = 35, find n.

3. If 2nC3 : nC

3 = 11:1, find n.

4. Verify that 8C4 + 8C

5 = 9C

4

5. Prove that 1

1

C

C

n

r

n

r

n

r when 1 .r n

6. If nCr-1

: nCr: nC

r +1 = 3 : 4 : 5, find r.


Example 1. A man has 6 friends. In how many ways can he invite one or more of

them to a party?

Sol. The different ways of combinations of inviting 6 friends are inviting exactly one,,

exactly 2, ........... exactly 6. These can be done in

6 6 6 6 6 6

1 2 3 4 5 6C , C , C , C , C , C ways.

The total number of ways in which one or more can be invited

= 6 6 6 6 6 6

1 2 3 4 5 6C C C C C C

= 6 6 6 6 6 6

1 2 3 2 1 0C C C C C C

= 6 + 15 + 20 + 15 + 6 + 1 = 63 ways.

Example 2. For a set of 5 true or false questions no student has written all correct

answers and no two students have given the same sequence of answers. What is

the maximum number of students in the class for this to be possible?

Sol. Each question can be answered in exactly 2 ways either T or F.

Total number of ways in which the 5 questions can be answered

= 2 × 2 × 2 × 2 × 2 = 32

Out of these 32 ways exactly one will be all correct answer.

Since no student has written all correct answers

The maximum possible number of students = 32 – 1 = 31.

Example 3. 4 friends shake hands mutually. Find the number of handshakes.

Sol. Let the friends be A, B, C and D. For one hand shake, there will

be 2 persons involved. To start with let person 'A' shakes hand

with the person 'B'. It is the same as person 'B' shaking hands

with person 'A'. AB and BA are the same.

So in finding the number of handshakes, we observe that order

is not important.

Hence the number of handshakes will be always

nC2, and nC

2 =

1

2

n n.

In this problem n = 4. Number of handshakes will be 4C2.

24

2

4! 4C

4 2 !2!

3 2 1

2 1 26

1

Example 4. Everybody in a function shakes hand with everybody else. The total

number of handshakes is 45. Find the number of persons in the function.

Sol. Let the number of persons in the function be n.

Then nC2 = 45

!45

2 !2!

n

n

145

2

n n

B

C

D

A

{AB, AC, AD}

C

D

B

{BC, BD}

C D

{CD}

86 UNIT-4

n(n – 1) = 90

n(n – 1) = 10 × 9

n(n – 1) = 10(10 – 1)

n = 10

Hence the number of persons in the function is 10.

Example 5. How many committees of five with a given chairperson can be selected

from 12 persons?

Sol. The chairperson can be choosen in 12 ways and following this the other four on

the committee can be choosen in 11C4 ways.

The possible number of such committees = 12 × 11C4

= 12 × 330 = 3960.

Example 6. There are 8 points such that any of 3 of them are non collinear. How

many straight lines can be drawn by joining these points?

Sol. A straight line is obtained by joining 2 points. Let the points be A and B.

The straight line is obtained by joining either A with B or B with A.

AB and BA are same.

This is a problem on combination.

Total number of lines that can be drawn out of 'n' non-collinear points = nC2

We know, 2

11

2

nC n n

Here, n = 8, r = 2

8C2 =

18 8 1

2 = 28

28 straight lines can be drawn by joining 8 points.

Example 7. There are 10 points such that any 3 of them are noncollinear. How many

triangles can be formed by joining these points?

Sol. A triangle is formed by joining 3 non-collinear points.

Here the order in which the points are joined is not at all important.

Total number of triangles that can be drawn out of 'n' non-collinear points = nC3

Here, n = 10, r = 3,!

! !

n

r

nC

n r r

10

3

10! 10!

10 3 !3! 7! 3!C

10 93

84

7!

7! 3 2 1 = 120

120 triangles can be formed by joining 3 points out of 10 non-collinear points.

A B

8

2

4

!

! !

8!

8 2 !2!

8

n

r

nC

n r r

C

Verification :

7 6!

6!8

2

2

28C


Alternate method:

3

!

! !

n nC

n r r

!

3 !3!

n

n

1 2 3 !n n n n

3 !n 3 2 1

3

1 2

6

n n n nC

If n = 10

10

3

10 10 1 10 2C

6

310 9 84

6 3

= 120

Example 8. How many diagonals can be drawn in a hexagon?

Sol. A hexagon has 6 vertices n = 6

A diagonal is obtained by joining the opposite vertices in pairs.

Total number of sides and diagonals = 6C2

2

1

2

n n nC 6

2

6 6 1

2C = 15

15 lines includes 6 sides. Number of diagonals = 15 – 6 = 9

Alternate method:

Number of diagonals = Number of straight lines formed – Number of sides of polygon

= nC2 – n =

11

2n n n

=

2 2

2

n n n 2 3

2

n n

3

2

n n

Number of diagonals in a polygon of n sides = 3

2

n n

In an hexagon, n = 6 Number of diagonals 3

2

n n

6 6 3

2 = 9

Example 9. The maximum number of diagonals in a polygon is 14. Find the number

of sides.

Sol. Number of diagonals in a polygon 3

2

n n

314

2

n n n(n – 3) = 14 × 2 n(n – 3) = 7 × 4

n(n – 3) = 7(7 – 4) n = 7

A

B C

88 UNIT-4

Alternate : 3

142

n n

n(n – 3) = 14 × 2 n2 – 3n – 28 = 0 n2 – 7n + 4n – 28 = 0

n(n – 7) + 4(n – 7) = 0 (n – 7) (n + 4) = 0

n – 7 = 0 OR n + 4 = 0

n = 7 n = – 4

'n' cannot be a negative number

Hence n = 7 (Heptagon)

Remember

When 'n' number of non-collinear points are given in a plane,

* number of straight lines = 2

1

2

n n nC * number of triangles = 3

1 2

6

n n n nC

* number of diagonals in a polygon = 2

3

2

n n nC n

Pascal Triangle

Pascal, Blaise (1623 - 1662) great French Mathematician , probabilist, combinatiorist,

physicist and philosopher.

Observe the triangle pattern discovered by Pascal.

Compare the corresponding elements in the two triangles.

The triangle on the right side is called Pascal triangle.

1C0

1C1

2C2

6C

6

5C5

3C3

4C4

6C5

5C4

4C3

3C2

2C1

1C0

2C0

3C1

4C2

5C3

6C4

3C0

4C1

5C2

6C3

4C0

5C1

6C2

6C1

5C0

6C0

1

1

1

1

1

6

5

4

3

2

1

1

3

6

10

15

1

4

10

20

1

5

156

1

1

1

1


EXERCISE 4.7

1. Out of 7 Consonants and 4 vowels, how many words of 3 consonants and 2 vowels

can be formed?

2. In how many ways can 5 sportsmen be selected from a group of 10?

3. In how many ways a cricket team of eleven be selected from 17 players in which

5 players are bowlers and the cricket team must include 2 bowlers?

4. How many (i) lines (ii) triangles can be drawn through 8 points on a circle?

5. How many diagonals can be drawn in a (i) decagon (ii) icosagon

6. A Polygon has 44 diagonals. Find the number of sides.

7. There are 3 white and 4 red roses in a garden. In how many ways can 4 flowers of

which 2 red be plucked?

8. In how many ways can a student choose 5 courses out of 9 courses, if 2 courses are

compulsory for every student?

9. There are 5 questions in a question paper. In how many ways can a boy solve one or

more questions?

10. In how many ways 4 cards from a pack of 52 playing cards can be chosen?

11. In an election there are 7 candidates and three are to be elected. A voter is allowed

to vote for any number of candidates not greater than the number to be elected. In

how many ways can we vote?

EXERCISE 4.8

1. If there are 6 periods in each working day of a school, in how many ways can one

arrange 5 subjects such that each subject is allowed at least one period?

2. A committee of 5 is to be formed out of 6 men and 4 ladies. In how many ways canthis be done when

(i) at least 2 ladies are included. (ii) at most 2 ladies are included.

3. A sports team of 11 students is to be constituted choosing at least 5 from class IXand at least 5 from class X. If there are 8 students in each of these classes, in howmany ways can the team be constituted?

4. In a vegetable mela or fair, an artist wants to make mascot (a toy that representsan organization) with Beans, Carrot, Peas and Tomato in a line.In how many ways can the artist make the mascot?

5. From a group of 12 students, 8 are to be chosen for an excursion. There are 3students who decide that either of them will join or none of them will join. In howmany ways can the 8 be chosen?

6. How many chords can be drawn through 20 points on a circle?

7. The English alphabet has 5 vowels and 21 consonants. How many words with 2different vowels and 2 different consonants can be formed from the alphabet?

8. In how many ways 5 letters be posted in 3 post boxes, if any number of letters can beposted in all of the three post boxes?

9. In a group of 15 students there are 6 scouts. In how many ways can 12 students be

selected, so as to include at least 4 scouts?

90 UNIT-4

ANSWERS

EXERCISE 4.1

1] 120 2] 12 3] 720 4] 336 5] 8 6] 20

EXERCISE 4.3

1] (i) 7! (ii) 18! (iii) 9!

5!(iv) 2 × 4! 2] (i) 720 (ii) 3,62,880 (iii) 40,200

(iv) 42 (v) 220 (vi) 870 3] (i) 210 (ii) 105 4] 720

5] (i) 3 (ii) 121 6] n(n + 1)

EXERCISE 4.4

1] (i) 11,880 (ii) 5,550 (iii) 40,320 (iv) 15 (v) 1

2] (i) 7 (ii) 15 (iii) 3 3] (i) 6 (ii) 8 4] (i) 5 (ii) 5

EXERCISE 4.5

1] 120 2] 2,880 3] 720 4] 1 5] 336 6] 90 7] 120

8] (a) 180 (b) 120 (c) 60 9] 210

EXERCISE 4.6

1] (i) 120 (ii) 1 (iii) 1,61,700 2] (i) 11 (ii) 7 3] 6 6] 27

EXERCISE 4.7

1] 25,200 2] 252 3] 2,200 4] (i) 28 (ii) 56 5] (i) 35 (ii) 170

6] 11 7] 18 8] 35 9] 31 10] 2,70,725 11] 63

EXERCISE 4.8

1] 3,600 2] (i) 186 (ii) 186 3] 3,136 4] 24 5] 117

6] 190 7] 50,400 8] 243 9] 435

Permutation and Combination

Fundamental

Principle of

Counting (FPC)

CombinationnC

r

PermutationnP

r

Factorial

notation n!

nPr =

!

( )!

nn r

where 0 r n

nCr =

!

!( )!

nr n r

If r = 0,nP

0 = 1

If r = n,nP

n = n!

If r = 1,nP

1 = n

If r = 0,nC

0 = 1

If r = n,nC

n = n

If r = 1,nC

1 = n

nPr = nC

r×r!

Documents

Class x English Maths Chapter04