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Derivation of the formula for nPr


Theorem
The number of permutations of n distinct elements is n!.

Proof
We can use the product rule as follows:

1st position: Any of the n elements.
2nd position: Any of the remaining (n - 1) elements.
3rd position: Any of the remaining (n - 2) elements.
4th position: Any of the remaining (n - 3) elements............(n - 1)th position: Any of the remaining two elements.
nth position: The one remaining element.

By Multiplying these n numbers together, we find the number of distinct permutations to be We now multiply and divide the right hand side by (n - r)! to obtain



Using the recurrence relation n! = n (n - 1)! repeatedly, we obtain
n (n - 1) (n - 2) ... (n - r + 1) (n - r)! = n!
Thus

Note that this formula works for r = n too, as 0! = 1.

What is the number of ways of arranging n elements taken none at a time?
If n distinct objects are lying before you and you want to arrange them taking none at a time, then you have just one way to do it. That is, let the things stay as they are. In other words, we have
nP0 = 1.
With this we see that the formula works for 0 r n.

We now list the values of nPr for some particular values of r.
nP0 = 1
nP1 = n
nP2 = n (n - 1)
nPn-1 =
and nPn = n!

 


Example
  1. Evaluate each of the following
    1. 10P3
    2. 9P5
Solution
  1. (a)
    (b)
  1. Find n if nP4 = 20 (nP2).
Solution


  1. n2 - 7n + 2n - 14 = 0 n(n - 7) + 2 (n - 7) = 0
    (n + 2) (n - 7) = 0 n = -2 or n = 7
    As n cannot be negative, n = 7.
  1. Find r if
    1. 5Pr = 6Pr-1
    2. 10Pr = 2. 9Pr
Solution
  1. We are given that 5Pr = 6Pr-1

    (7 - r) (6 - r) = 6
    42 - 13 r + r2 = 6
    r2 - 13r + 36 = 0 (r - 4) (r - 9) = 0 r = 4, 9.
    But r cannot exceed 5, therefore, r = 4.
  2. We are given that 10Pr = 2. 9Pr

    10 = 2 (10 - r)
    10 = 20 - 2r
    2r = 20 - 10 = 10 or r = 5.
  1. 9P5 + 5 (9P4) = 10Pr, find the value of r.

Solution

  1. We are given that 9P5 + 5 (9P4) = 10Pr
     




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