# Permutations

## Permutations:

The word permutation means arrangements. Each of the different arrangements that can be made by taking some or all of a given set of objects at a time is known as their permutation.

Thus, considering the letters A, B, and C of the English alphabet, the different permutations are as follows:

• by taking 1 at a time: A, B, C
• by taking 2 at a time: AB, BA, BC, CB, AC, CA
• by taking all at a time: ABC, ACB, BCA, BAC, CAB, CBA

The number of permutations of n things taken r at a time is noted by nPr or P(n, r). In the symbol nPr or P(n, r) both n and r are positive integers and n ≥ r.

### Number of Permutations of n different things by taking r(n ≥ r) at a time:

Consider that n different things are n differently colored marbles and let r(r n) vacant holes are to be filled up by taking r marbles. So, the number of permutations of n different things taken r at a time will be the same as the number of ways in which the r holes can be filled up by taking r marbles from the set of n differently colored marbles at our disposal.

Now, any one of the n marbles can be placed in the first hole concerned. So, the first hole can be filled up in n different ways.

When the first hole has been filled up in any one of the n ways, there will be (n – 1) marbles and any one of them can be placed in the second hole. So, the second hole can be filled up in (n – 1) ways. As each way of filling up the first hole can be associated with each way of filling up the second, by the fundamental principle of multiplication, the first two holes can be filled up in n (n – 1) ways.

Now, when the first two holes have been filled up, the third hole can be filled up by any one of the remaining (n – 2) marbles and so the third hole can be filled up in (n – 2) different ways. Since each of these (n – 2) ways of filling up the third hole can be associated with the n (n – 1) ways of filling up the first two holes, the first three holes can be filled up in n (n – 1) (n – 2) ways.

So far three holes have been filled up and we find that the result contains as many factors as the number of holes filled up. Also, it is clear that as we proceed to fill up a new hole, the number of factors goes on increasing by a new factor and each new is one less than the previous factor.

Thus, the number of ways in which r holes can be filled up = n (n – 1) (n – 2) … to r factors. Here, the rth factor, n – (r – 1) = n – r + 1.

Hence, the required number of permutations,