Question

(n-1)!/(n-r-1)^+r.(n-1)!/(n-r)!=n!/(n-r)!

Answer 1

Answer:

The permutation of n things taken r at a time is defined as the number of ways in which you can arrange r objects chosen from n distinct objects in a queue. Mathematically, it is denoted as n at the top left corner and r at the bottom right corner of P (I’ve seen P(n,r) being used as a common form of notation too).

Using the fundamental principle of counting, one can arrive at the much widely known closed expression for P(n,r):

Suppose there are r places. You can fill the first place by placing picking any of the n objects and placing it in the first place. This can be done in n ways. So, the first place has been filled up and we have n−1 objects left. Similarly, the second place can be filled by picking any of the remaining n−1 objects and placing it in the second place, and now we have n−2 objects left. This process will go on till the rth place has been filled up. Now, we happen to make a small observation:

1st place: n=n−(1–1) ways

2nd place: n−1=n−(2–1) ways

3rd place: n−2=n−(3–1) ways

This heavily suggests that the rth place can be filled up in n−(r−1)=n−r+1 ways.

Now, since the events are occurring simultaneously, the total no. of possible arrangements is simply the product of the number of ways of filling each of the r places.

Hence, the required expression is n(n−1)(n−2)…(n−r+1), as you’ve put up in your question.

Now, in order to close the expression, multiply and divide the expression by (n−r)!. Clearly, the numerator becomes n!. Hence, the closed expression for P(n,r) is n!(n−r)!.