Permutation Theorem I – Proof and Example | Mathematics Class 12

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Permutation MathematicsMathematics | Notes
Permutation Theorem – 1 | Proof and Example
For: Science Class : 12

Permutation is the arrangement of objects with respect to some order. The permutation on n objects taken r at a time is denoted by P(n, r) or nPr.

Theorem-1
The total no of permutation of a set of n objects taken r at a time is given by:
P(n, r) = n.(n – 1).(n – 2).(n – 3) ………….. (n – r + 1)
[OR]
P(n, r) = n!/(n – r)! [where : ! is sign for factorial]

Proof:
The total number of permutation of a set on n objects taken r at a time is equivalent to filling of those n objects in r positions.

  • To fill up the first position there are n choices. So, the first position can be filled up by n ways.
  • For the second position (n – 1) objects left there to fill up. So, second position can be filled up by (n – 1) ways.
  • Similarly, the third position can be filled up by (n – 2) ways.
  • Continuing this process up to r^th position so, the r^th position can be filled up by {n – (r – 1)} ways.
  • Hence, the total number of permutation of set on n objects taken r at a time is

n. (n -1).(n – 2)………{n – (r – 1)} ways
I.e. P(n, r) = n.(n -1).(n – 2)………{n – (r – 1)}
or, P(n, r) = [n.(n -1).(n – 2)………{n – (r – 1)}.(n – r).{(n – (r + 1)}……..3, 2, 1] / [(n – r). {n – (r + 1)}….. 3, 2, 1]
or, P(n, r) = n! / (n – r)!

Note:
P(n, n) = n! / (n – n)! = n! / 0! = n! / 1 [Because 0! = 1] = n!
i.e. P(n, n) = n!

Permutation of object not all different
The number of permutation of a set of n objects taking all at a time when P objects are of 1st kind, Q objects are of 2nd kind, and R objects are of 3rd kind is given by n! / (P! Q! R!)

For Example:
Q.1. How many ways the letters of word ‘CALCULUS’ can be arranged.

CALCULUS

Total number of objects = 8
Number of C’s = 2
Number of L’s = 2
Number of U’s = 2

Therefore, total Number of permutation = 8!/ (2! 2! 2!)
= [8 × 7 × 6 × 5 × 4 × 3 × 2 × 1] / [2 × 1 × 2 × 1 × 2 × 1]
= 5040

Posted By : MeroSpark | Comment RSS | Category : Class XI, HSEB Notes
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One Comment

  1. Posted March 3, 2015 at 5:02 pm

    Quite interesting. I really understood it clearly. Thanks for this. I hope to get all the old questions and all the chapters’ brief and easy explaination.

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