Math StatisticCambridge University Press

By: danica p

Permutation and

Combination

Permutation, what s it?Permutation is an arrangement of elements of a set in which the order of the elements considered.Means, it explain the number of ways in which a subset of objects can be selected from a given set of objects, where order is important.

Example: Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is important? Answer: (AB, AC, BC, BA, CA, CB) there is 6 arrangements are possible.

from A: A-B, A-C (2 arrangements)From B:B-A, B-C (2 arrangements)From C:C-A, C-B (2 arrangements) Total: 2*3 = 6 arrangements

Notice!PERMUTATION (P)

= ARRANGEMENT !

Factorial Formula of Permutation

!.

( )!n rn

Pn r

Notation of n = n!

n! = n*(n-1)*(n-2)*.......*2*1e.g. 9!=9*8*7*6*5*4*3*2*1

* = multiply

Example:15 runners are hoping to take part in a marathon competition, but the track has only 4 lines. How many ways can 4 of the 15 runers be assigned to lanes?Answer: Using Factorial formula of P,15P4 = = 32760

1*2*3*4

1*2*3*4*5*6*7*8*9*10*11*12*13*14*15

)!1115(

!15

You can using your scientific calcuator, it simple, you enter number n, then shift-x-then enter the number of r, try it! 15-shift-x-P- then, 4, so it became to 15P4 = 32760

Then, what about a combination?Combination is an arrangement of elements of a set in which the order of elements is not considered. Means: The number of ways in which a subset of objects can be selected from a given set of objects, where order is not important.

Example:Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is not important? Answer: (AB, AC, BC) 3 combinations.

each alphabet has different couple combination, and that is:A can together with B,A can together with C,B can together with C, total: 3

Notice!COMBINATION (C)

= CHOOSE!

Factorial Formula for Combinations

!.

! !( )!n r

n rP n

Cr r n r

Example:The manager of a football team has a squad of 18 players. He needs to choose 11 to play in a match. How many possible teams can be choosen?Answer: Using Factional formula of C,18C11= = 31824

!7!*11

!18

)!1118(!11

!18

!11

1118

P

You can using your scientific calcuator, it simple, you enter number n, then shift-()-then enter the number of r, try it! 18-shift-(÷)-P- then, 11 so it became to 18C11= 31824

Difference between Permutation and Combination

Exercises:

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.756 E.None of theseAnswer: Option DExplanation: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 5x6 x 5+ (7C3 x 6C1) + (7C2)3 x 2 x 12 x 1= 525 +7 x 6 x 5x 6+7 x 63 x 2 x 12 x 1= (525 + 210 + 21)= 756.

2. 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 theseAnswer: Option CExplanation: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 x6 x 5+3 x 2x 6+ 12 x 12 x 1= (45 + 18 + 1)= 64.

3. 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.5040E.None of theseAnswer: Option CExplanation: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 (4 + 1 = 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.

4. In how many ways can the letters of the word 'LEADER' be arranged?A.72 B.144 C.360 D.720 E.None of theseAnswer: Option CExplanation:The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R. Required number of ways = = 360

)!1)(!1)(!1)(!2)(!1(

!6

Thank you!!