Permutations

The different arrangements that can be made with a given number of things taking some or all of them at a time are called permutations.

The symbol ⁿP_r or P(n,r) is used to denote the number of permutations of n things taken r at a time.

Examples:

1. 2 and 3 are two digits and with these digits, the numbers 32 and 23 are formed. Although both numbers viz., 32 and 23 consist of the digits 2 and 3, the order of digits is different. Each of the above arrangements is called a 'permutation'. Thus, the number of arrangements or permutations of two distinct digits 2 and 3 is 2.

2. The permutation of the three letters a, b, c taken two at a time are

Theorem:

The number of permutations of n dissimilar things taken r at

Proof:

The number of permutations of n different things taken r at a time is the same as the number of ways of filling n letters and r blank spaces, supposed to be arranged in a straight line as shown above. Each blank is accommodating only one letter. We may fill the first blank with any one of the n letters. Therefore, the first blank can be filled in n different ways. Having filled the first blank in any one of these ways, we have

(n-1) letters with which to fill the next blank. Hence, the second blank can be filled in (n-1) ways, after filling the first blank in n different ways.

Therefore, there are n(n-1) different ways of filling the first two blank spaces.

Having filled the first two blank spaces in any one of these ways, we have (n-2) letters with which to fill the third blank space, for each of the n(n-1) ways of filling the first two.

There are n(n-1)(n-2) different ways of filling the first three blank spaces. Proceeding in this way, we see that there are n(n-1)(n-2)(n-3)(n-4)....[r factors] different ways of filling r blank spaces with n letters. The r^th factor = n - (r -1) = n - r = 1.

The number of r-permutations of n different things is ⁿP_r = P(n,r) = n(n-1) (n-2) (n-3)...(n - r +1)

Corollary 1:

If we put r = n in the above formula, then

Note:

Strictly speaking 0! has no meaning. But since ⁿP_n = n! we may understand 0! = 1.

Corollary 2:

Corollary 3:

Corollary 4:

Proof:

First method:

P(n,r) is the number of permutations of n dissimilar things taken r at a time. These permutations can be divided into two groups.

(i) Those not containing a particular thing l.

(ii) Those containing a particular thing l.

Taking out l from the given things, we have (n-1) things which can be arranged in r blanks in P(n-1,r) ways. Then, the number of permutations not containing l is P(n-1,r).

Next in the permutations containing l, it occupies one of the r blank spaces in r ways. The remaining (r-1) blank spaces can be filled with the remaining (n-1) things in P(n-1,r-1) ways. Thus, the number of permutations containing l is lP(n-1, r-1).

Hence, the total number of permutations is P(n-1,r) + rP (n-1, r-1).

Second method: