Algebra - permutation

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Algebra – 3rd

secondary Permutation - 1 -

Permutation

Permutation and Combination Introduction: Both are ways to count the possibilities, the difference between them is whether

order matters or not for example: consider Ahmed, Mohamed and Amr

Is that the same as Mohamed, Ahmed and Amr? Does the order of the names matter?

If yes, then we are dealing with permutations

If no, then we are dealing with combinations

A permutation is " a re-arrangement of elements of a set " . يعتمد على الترتيب With permutations, every little detail matters : Ahmed , Amr and Amir is different from Amir,

Amr and Ahmed “ for example , if the three are in champion ship “

So, what does this mean? It means a permutation is ONLY interested in re-arranging the

elements of the set... Any pictures of the collected elements in different orders is fine.

A permutation therefore tends to be a large number.

Simple Example

Taking the 4 letters, ABCD, write down all the permutations of 3 of these letters:

ABC BAC CAB DAB

ACB BCA CBA DBA

ABD BAD CAD DAC

ADB BDA CDA DCA

ACD BCD CBD DBC

ADC BDC CDB DCB

Here, if you like, the order matters, since ABC is different to ACB and different to BCA and

different to CAB …… etc. , Permutations see these as all different answers.

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Question How many ways can we pick a Gold, Silver, and Bronze medal in a championship which

consists of 8 men .

Answer

We’re going to use permutations since the order we hand out these medals matters. Here’s

how it breaks down: Gold medal : 8 choices: A B C D E F G H ……… Let’s say A wins the Gold.

Silver medal : 7 choices: B C D E F G H ……… Let’s say B wins the silver.

Bronze medal : 6 choices: C D E F G H ……… Let’s say C wins the bronze.

We picked certain people to win, but the details matter: we had 8 choices at first , then 7 then

6 , So The total number of options was 8 × 7 × 6 = 336.



Algebra – 3rd


Combination

Combinations on the other hand, are pretty easy going. The details don’t matter. Ahmed , Amr and Amir is the same as Amir, Amr and Ahmed .

A combination is "one or more elements selected from a set without regard to the order"

The "without regard" means that the collection matters rather than order in combinations, so

in the above example, the fact we ABC, ACB, BAC, BCA, CAB, CBA... for combinations, these

are all 1 combination of letters A, B and C.

So, questions concerning picking a team of 5 people from a squad of 11... you would need

combinations, not the fact that you have so many different permutations of these 5 people.

Example

Taking the 4 letters, ABCD, write down all the combinations of 3 of these letters:

ABC ABD ACD BCD → “ each number didn’t repeat more than once”

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Question

How many ways can I give 3 tin cans to 8 people?

Answer

Well, in this case, the order we pick people doesn’t matter. If I give a can to Alice, Bob and

then Charlie, it’s the same as giving to Charlie, Alice and then Bob. Either way, they’re going to be equally disappointed.

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Some examples

Picking a team of 3 people from a group of 10 → Combination

Picking a President, VP and Waterboy from a group of 10 → Permutation

Choosing 3 desserts from a menu of 10 → Combination

Listing your 3 favorite desserts, in order, from a menu of 10 → Permutation

So Permutations are for lists (order matters) and combinations are for groups (order doesn’t matter).

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Algebra – 3rd


Permutation

n

rP , r n , n Z

The first seat could be occupied by 4 ways A or B or C or D

The second seat could be occupied by 3 ways B or C or D

The third seat could be occupied by 2 ways C or D

The number of ways of sitting in 3 adjacent seats = 4 3 2 =24

General formula =

Example : In how many different forms can 4 students A , B , C , D sit in three adjacent seats ?

Answer

Any of the students A or B or C or D can sit on the first place or second or third or fourth

Such as :

And So on …………….

This could be written using the general formula as : 4

3P 4 3 2 =24

Where 4 “n” is a number of students needed to be arranged in 3 different ways “r”

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Example :How many ways can we pick a Gold, Silver, and Bronze medal in a championship

which consists of 8 men .

Answer

As we said before , We picked certain people to win, we had 8 choices at first , then 7 then 6 , So The total number of options was 8 × 7 × 6 = 336.

This could be written using the general formula as : 8

3P 8 7 6 =336

A B C

A C B

A C D

A B D



Algebra – 3rd


! Or

! Or

n

r

n n!P

n r n r !

7

2

12

4

n 1

3

n 1

4

7 7 7 6 5 4 3 2 11) P 7 6

7 2 5 5 4 3 2 1

12 12 12 11 10 9 8 7 6 5 4 3 2 12) P 12 11 10 9

12 4 8 8 7 6 5 4 3 2 1

n 1 n 2 n 3 n 4 n 5 ........ 3 2 1n 13) P n 1 n 2 n 3

n 4 n 4 n 5 ........ 3 2 1

n 14 ) P

n 3

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

n 1 n n 1 n 2n 3 n 4 ........ 3 2 1

7

2

12

4

n 1

3

n 1

4

1) P 7 6

2) P 12 11 10 9

3) P n 1 n 2 n 3

4 ) P n 1 n n 1 n 2

In General :

n

rP n n 1 n 2 n 3 ............. n r 1

For example :

So , it is clear that : n

rP is the product of r positive integers , these integers begin with n ,

and each integer is less than the previous by 1 .

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Factorial :

If each of the following integers goes in order until it reaches 1 , Then this is called Factorial

And it is denoted by :

For example : factorial (4) = 4 4! 4 3 2 1

factorial (8) = 8 8! 8 7 6 5 4 3 2 1

factorial (2) = 2 2! 2 1

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Another counting method rule on Permutation



Algebra – 3rd


6

6

50

50

6 P 6 5 4 3 2 1

Also

50 P 50 49 48 ..... 3 2 1

10 10 9 10 9 8 10 9 8 7 ........ So on

29 29 28 29 28 27 ....

n 1 ( n 1) n ( n 1)n n 1 ....

10 10 910

9 9

15 15 14 1315 14 210

13 13

n 1 ( n 1)n( n 1) n 2( n 1)n( n 1)

n 2 n 2

n n n 1

proof : n n( n 1)( n 2 ) .... 3 2 1 n ( n 1)( n 2 ).... 3 2 1 n n 1

n

r

nP

n r

Very important remarks

1) n

nn P n( n 1)( n 2 ) .......... 6 5 4 3 2 1

for example :

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2) n n

0 0

n nP 1 Proof : P 1

n 0 n

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3) n

n

n n n0 1 Proof : P n 0 1

0 0 n

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4)

for example :

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5) Use first law of permutation :

n

rP n n 1 n 2 n 3 ............. n r 1 if r is small number

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6) But you can Use the second law of permutation :

if r is big or unknown

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Algebra – 3rd


29 50 n 1Simplify the following : a) b) c)

30 48 n 2

2 3

29 29 1a )

30 30 29 30

50 50 49 48b ) 50 49 2450

48 48

n 1 n n 1 n 2n 1 c) n 1 n n 1 n n 1 n n

n 2 n 2

n 1if 72 , Then find the value of n

n 3

2

2 2

n 1 n 2 n 372 n 1 n 2 72 n 2n n 2 72

n 3

n 3n 2 72 0 n 3n 70 0 n 10 n 7 =0

n 10 " agreed" n 7 " refused as n Z "

2 3 4 1 1 1L.H.S : Tooo long

6 5 4 3 2 7 6 5 4 3 8 7 6 5 4 360 840 1680

L.H .S. : multiply all numerator and denominator by the highest Factorial which is 8

8 2 8 3 8 4 8 7 6 2 8 7 3 4 8 7 2 8 3 4

8 6 8 7 8 8 8 6 8 7 8 8 8 8

56 2 8 3 4 112 48 24 184 23L.H .S R.H .S

8 8 7 8 7 7

Examples

Example (1)

Answer

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Example (2)

Answer

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Example (3)

2 3 4 23Prove that :

6 7 8 7

Answer

1st method : off course the first way which will come to your mind will be :

2nd

method : This way which you must use :



Algebra – 3rd


2 2

n 8

2 2

3 n 13 5 208 5 208

n 2 n 1 n 2 n n 1 n 2 n 1 n 2 n 1 n 2 n n 1 n 2

3 n 1 5 208 208 2083n 3 5 3n 2

n 1 n 2 n n 1 n 2 n n

3n 2n 208 3n 2n 208 0 3n 26 n 8

26n " refused" Or n 8 " agreed"

3

P P 8 7 56

Example (4)

n

2

3 5 208If , Then find P

n 2 n 1 n

Answer

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Example (5)

If 2n n

3 2P 20 P , Then find the value of n .

Answer

2n 2n 1 2n 2 20 n n 1 2 2n 2n 1 n 1 20 n n 1

204n 2n 1 20 n 4 2n 1 20 2n 1 5

4

2n 6 n 3

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Example (6) 2n 1 2n 1

n 1 nIf P : P 3 : 5 , Then find n 2

Answer

2 2

2n 1 2n 1 3 2n 1 n 1 3

2n 1 n 1 2n 1 n 5 n 2 2n 1 5

2n 1 2n 2n 1 n 1 3 2( 2n 1) 3

n 2 n 1 n n 1 2n 1 5 ( n 2 )( n 1) 5

3n 9n 6 20n 10 3n 11n 8 0

1( 3n 1)( n 4 ) 0 n 4 and n " refused"

3

n 2 4 2 6 720

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Algebra – 3rd


9 9

r 3

2r 1 7

4 4

9r 504 P 9 8 7 P

r 3 P P 7 6 5 4 840

P

2r 1 9

3 3

13 13r r17160 13 12 11 10

r 4 P P 9 8 7 504

P P

4

n

4If P 840 7 6 5 4 Then n 7

2

2

1

77

142

2 720

2 360

2 180

3 90

3 30

2 10

5 5

1

720 1 2 3 4 5 6 n 6 5 4 3 2 1 6

n 6 and n 2 4 4 3 2 1 24

Example (7) 9 2r 1

r 4If P 504 , Then find P

Answer

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Example (8)

13 2r 1

r 3If P 17160 , Then find r and P

Answer

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Example (9)

n

4If P 840 , find n

Answer

5 840

3 168

56

28

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Example (10)

If n 720 , then find n and n 2

Answer

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Algebra – 3rd


n 10

3 3

1120 1 2 3 4 5 n 120 5 4 3 2 1 5

2

1n 5 n 10 P P 10 9 8 720

2

20

20

40 40 39 38 37 .... 4 3 2 1 40 38 36 ..... 4 2 39 37 35 ..... 5 3 1

40 2 20 2 19 2 18 ..... 2 2 2 1 39 37 .... 3 1

2 20 19 18 ..... 3 2 1 39 37 .... 3 1

2 20 1 3 5 ..... 37 39

50

50

100 100 99 98 97 .... 4 3 2 1

100 98 96 ..... 4 2 99 97 95 ..... 5 3 1

100 2 50 2 49 2 48 ..... 2 2 2 1 99 97 95 ...... 3 1

2 50 49 48 ..... 3 2 1 99 97 95 ...... 3 1

2 50 1 3 5 ..... 97 99

Example (11)

n

3

1If n 120 , Then find P

2

Answer

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Example (12)

n

oIf 15 P x 15 ,Then find x 14

Answer

n

oP 1 15 x 15 15 x 15 14 x 14 x 14 0 1

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Example (13)

20Prove that: 40 2 20 (1 3 5..... 39 )

Answer

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Example (14)

50Prove that : 100 2 50 1 3 5 ..... 99

Answer

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Algebra – 3rd


n r 1 nn n r n

r r 1 n r r 1 n r 1 n r r r 1 n r n r 1 r r 1 n r 1 n r

n r 1 n r n n n r 1 r n 1 n n 1R.H .S

r r 1 n r 1 n r r n r 1 r n r 1 r n r 1

6 6

r r 1

6 4 6 1 4 1 4P 4 P

6 r 6 r 1 6 r 7 r 6 r (7 r ) 6 r

7 r 4 r 3

r 1 r 2 r 1 4 5 2 4 3 5 2 1 1 7L.H .S. 4 R.H .S

r 2 r 3 r 3 6 3 3 6 5 3 2 6 3 10

x y 9

3 3

x y 5

4 4

x 7

y 2

P 504 9 8 7 P x y 9.....(1)

P 120 5 4 3 2 P x y 5.....( 2 )

from(1),( 2 ) x 7 And y 2 , Also P P 7 6 42

Example (15)

n n n 1Provethat :

r n r r 1 n r 1 r n r 1

Answer

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Example (16) n n 1 n 1

r r r 1Prove that : P P r P

Answer

n

r

n 1 r n 1 n 1 r n 1

n 1 r n r n 1 r n r n r 1

n 1 n r rn r n 1 n r n 1 r n 1r n 1

n r n 1 r n r n r 1 n r n 1 r n r n 1 r

n n 1 nP

n r n r 1 n r

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Example (17)

6 6

r r 1

r 1 r 2 r 1 7If P 4 P , Then Prove that :

r 2 r 3 r 10

Answer

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Example (18) x y x y x

3 4 yIf P 504 and P 120 , Then find P

Answer



Algebra – 3rd


4

8

8

4

Don't ever write P because 4 < 8

So the right Number of methods = P 8 7 6 5 1680

5 60480

3 12096

3 4032

3 1344

2 448

2 224

2 112

2 56

2 28

2 14

7 7

1

n

6

n

6

r 720 1 2 3 4 5 6 6 r 6

P 60480

P 60480 9 8 7 6 5 4 n 9

12

4X 4, 3, 2, 1,0,1,2.....,7 n( X ) 12 n(Y ) P 12 11 10 9 11880

Example (19) n n 2

4 3If P 14 P , Then find 3n 24

Answer

n n 2

4 3

2

n 14 n 2 n( n 1) n 2 14 n 2P 14 P

n 4 n 2 3 ( n 4 ) n 5 n 5

n n 1 14 n 4 n 15n 56 0 ( n 7 )( n 8 ) 0

n 7 or n 8

When n 7 : 3n 24 is refused

When n 8 : 3n 24 0 1

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Example (20) n

rIf P 60480 and r 720 ,Then find n and r

Answer

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Example (21)

Let X { x : x Z , 5 x 7 } and Y {( a,b,c,d ) : a,b,c,d X ,a b c d }, then find n(Y ).

( Note : n Y means find the numbers in the Y element )

Answer

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Example (22)

By how many ways is it possible to 4 boys to sit in 8 seats arranged in a raw?

Answer



Algebra – 3rd


How many different numbers can be formed out of the digits 3,4,5,6,7,8,9.

i) If each number contains 3 different digits .

ii) If each number contains all these digits without repeating .

iii) If each number contains 5 different digits and can be divided by 2 .

iv) If each number contains 4 different digits and starts with the digit 7 .

v) If each number contains 4 different digits and is less than 6000 .

we can put any one of

the 7 numbers

So , we have 7 ways

6 numbers remain

So , we have 6 ways

5 numbers remain

So , we have 5 ways

Find the number of permutations can be formed if we take letters from the word " Modern "

in each of the following :

(a) each permutation contains only 3 letters .

(b) each permutation contains only 6 letters .

( c ) each permutation starts with " M " and ends with " N "

1 way 1 way2 ways3 ways4 ways

So , we have 1 way

to start with M

So , we have 1 way

to ends with N Now , the remaining

numbers are 4

" O , D , E , R "

1 way

Example (23)

Answer

6

3

6

6

(a) number of permutations P 6 5 4 120

(b) number of permutations P 6 6 5 4 3 2 1 720

(c)

number of permutations 1 4 3 2 1 1 24 1 1 4 24

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Example (24)

Answer

i)

7

3P 7 6 5 210 numbers

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ii) which means that the 7 numbers are distributed to 7 places 7

7P 7 7 6 5 4 3 2 1 5040

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Algebra – 3rd


3 ways4 ways5 ways6 ways

start from here

we have 3 ways

" either 4 or 6 or 8 "

3 numbers remain

So , we have 3 ways

4 numbers remain

So , we have 4 ways

6 numbers remain

So , we have 6 ways

5 numbers remain

So , we have 5 ways

1 way to put 7 6 numbers remain

So , we have 6 ways

5 numbers remain

So , we have 5 ways

4 numbers remain

So , we have 4 ways

1 way 6 ways 5 ways 4 ways

let 3 is put , then

6 numbers will

remain

3 ways 6 ways 5 ways 4 ways

3 ways

iii) we have 4 , 6 and 8 which are divided by two , if we take 4 in the last list , then the rest

numbers “ 3 , 5 , 6 , 7 , 8 , 9 “ will be distributed to the remaining lists

The number of numbers 6 5 4 3 3 1080

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iv)

The number of numbers 1 6 5 4 120

v) if the numbers start with 3 or 4 or 5 , these will be less than 6000

The number of numbers 3 6 5 4 360

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Example (25)

If X = 3,4,5,6,7 , find without repeating any digit , each of the following :

a) How many 5 digits number can be formed from element X.

b) How many 5 digits number can be formed from element X such that

the unit digit is

neither 4 nor 5

c) How many 5 digits number can be formed from element X such that the unit digit is

not 4 and the tens digit is not 5

Answer

a) 5

5The number is P 5 5 4 3 2 1 120



Algebra – 3rd


2 ways1 way3 ways4 ways

start from here

we have 2 ways

" either 4 or 5 "

2 ways

1 way1 way2 ways3 ways

start from here we

have only 1 way

to get 4

1 way

start from here we

have only 1 way

to get 5

the rest numbers are

either 3 or 6 or 7

b) let’s see the number of ways if the unit digit is 4 or 5 :

if we choose 4 in the last list , then the remaining numbers

will be “ 3 , 5 , 6 , 7 “

So the number of ways in which 4 and 5 appear in the unit digit are : 4 × 3 × 2 × 1 × 2 = 48

Then the number of digits such that 4 and 5 are excluded from the unit digits are :

120 – 48 = 72 ways

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c) let’s see the number of ways if the unit digit is 4 and the tense digit is 5 :

So the number of ways in which 4 in the unit digit and 5 appear in the tense digit are :

3 × 2 × 1 × 1 × 1 = 6

Then the number of digits such that 4 is excluded from the unit digit are and 5 is excluded

from the tense digit are :

120 – 6 = 114 ways

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Algebra - permutation