CBSE & NCERT Solutions For Class 11 Maths Permutation
A Permutation is an arrangement of objects from a certain set. For example, ABC, ADB and
BAC are permutations of 3 letters taken from the letters ABCDE.
Two principles underlie even the most sophisticated counting methods.
Suppose that a choice can be made from options and another choice can be made from
options. Then
- The number of ways to make either the first choice or the second choice is assuming that all the options are different (the addition principle).
- The number of ways of making the first choice and the second choice is assuming that all combinations of the two choices are allowed (the multiplication principle).
Example: Calculate P (10, 3), the number of photographs of 10 friends taken 3 at a time.
P (10, 3) = 10 · 9 · 8 = 720.
Note that you start with 10 and multiply 3 numbers.
A general formula, using the multiplication principle:
P (n, k) = n · (n − 1) · (n − 2)· · ·(n − k + 1).
Note that there are k consecutive numbers on the right hand side.
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Example: In how many ways can you choose a President, secretary and treasurer for a club from 12 candidates, if each candidate is eligible for each position, but no candidate can hold 2 positions? Why are conditions 1, 2and 3 satisfied here?
P (12, 3) = 12 × 11 × 10 = 1, 320.
Condition 1 is satisfied because we have a single set of 12 for all 3 positions.
Condition 2 is satisfied because no one can hold more than one position.
Condition 3 is satisfied because being president is different than being treasurer or secretary.
Example: How many words can be made from rearrangements of the word BANANA?
{B, A, N, A, N, A} = {A, B, N}.
The ‘A’ is repeated 3 times.
The ‘N’ is repeated 2 times.
The ‘B’ is repeated once.
Hence the answer is 6!
1! · 2! · 3! = 60.
Example: How many words can be made from rearrangements of the letters of the word BOOKKEEPER?
10!
1! · 3! · 2! · 2! · 1! · 1! = 151, 200.
There are 10 letters in BOOKKEEPER. In alphabetical order,
B ↔ 1, E ↔ 3, K ↔ 2, O ↔ 2, P ↔ 1, R ↔ 1.
Note that the total number of letters is the sum of the multiplicities of the distinct letters: 10=1+3+2+2+1+1.
Example: A family goes to a studio to have their photo taken, and the photographer must arrange them to get the best shot. Three of them have red shirts, four of them have blue shirts, and two of them have black shirts. How many ways can the photographer arrange the family members?
In this example, there are 9 family members total, regardless of shirt color, so the n value is 9. In the denominator the 3 family members wearing red would be the n1 value, the 4 family members wearing blue would be the n2 value, and the 2 family members wearing black shirts would be the n3 value.
P=9! / 3! *4! *2!
In the numerator, 9 factorial is 362,880 in the numerator, and 3! 4! and 5! multiplied
together in the denominator equal 288.
P = 9! / 3! *4! *2!
= 9*8*7*6*5*4*3*2*1/3*2*1*4*3*2*1*2*1.
Simplify the fraction by dividing.
There are 1,260 ways to arrange the nine family members by shirt color.
Example: A family goes to a studio to have their photo taken, and the photographer must arrange them to get the best shot. Three of them are male and five of them are female. How many ways can the photographer arrange the family members?
In this example, there are a total of 8 family members regardless of gender, so the n
value is 8. Either group of identical objects can be substituted as the x value because
the denominator comes out the same no matter what. Notice that, whatever the case,
the sum of the numbers in the denominator (5 and 3) is always equal to the number
If men are the x value:
P= 8! / (8! -3!)3!
=8! /5! *3!
If women are the x value:
P=8! / (8-5)! *5!
= 8! /3! *5!
In the numerator, 8! equals 40,320 while in the denominator 3! multiplied by 5!
equals 720.
P=8! /5! *3! = 8*7*6*5*4*3*2*1 / 5*4*3*2*1
= 40320/720
= 56
There are 56 ways the photographer can arrange 3 men and 5 women.
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