    # Different Formulas On Permutation Theorem 1:

To find the number of permutations of n different things taken r at a time (r ≤ n).

The number of permutations of n different things, taken r at a time, none of things being repeated in any permutation. Suppose, we have r blank spaces and n ( ≥ r) different things. Then the required number of permutations will be obviously the same as the number of ways in which r blank places can be filled up with n different things. Clearly, the first place can be filled up in n different ways, for any one of the n different things can be placed there. So when the first place has been filled up in any one of these n ways, the second place can be filled up in (n-1) ways, for any one of the remaining (n - 1) things can be taken to fill up the second place. Clearly, each way of filling up the first place can be associated with each of the (n - 1) ways of filling up the second place. Therefore, the first two places can be filled up in n (n - 1) ways. Again, when the first two places have been filled up in any one of these n (n - 1) ways, the third place can be filled in (n - 2) ways, for any one of the remaining (n - 2) things can be placed there. As each way of filling up the first two places can be associated with each of the (n - 2) ways of filling up the third place, the first, second and third places can be filled up in n (n - 1) (n - 2) ways.

Continuing similarly and noting that a new factor is introduced with each of the new place filled up and the number of factors at any stage is equal to the number of places filled up, we can say that the number of ways in which r blank places can be filled up by n different things is

n (n - 1) (n - 2) ..... to r factors

where rth factor = n - (r - 1) = n - r + 1 [Since the second factor = n - 1 = n - (2 - 1); the third factor = n - 2 = n - (3 - 1) and so on].

Therefore, the required number of permutations of n different things taken r at a time

= n (n - 1) (n - 2) ........ (n - r + 1).                 ..... (1)

Corollary: The number of permutations of n different things taken all at a time (or permutations of n different things among themselves)

= n (n - 1) (n - 2) ........ (n - n + 1) [Using r = n in (1)]

= n (n - 1) (n - 2) ........3 ⋅ 2 ⋅ 1                       ......(2)

Notations: Theorem 2:

To find the number of permutations of n things taken them all at a time, when p of the things are alike of one kind, q of the things are alike of another kind, r of the things are alike of a third kind and the rest are all different.

We denote the n things by n letters and assume that p of them to be a, q of them to be b, r of them to be c and the rest to be different.

Let consider x to denote the required number of permutations.

Obviously, there are p a's in each of the x permutations. Now, let us consider any one of these x permutations and replace the p a's by p new letters a 1, a2 ......ap different from each other and also different from each other and also different from the remaining (n - p) letters. Clearly these new p letters can be arranged among themselves in p! ways. Therefore, if this change is made in each of the x permutations, we will get x * p! permutations in which p a's are will be all different.

Let us consider any one of these x * p! permutations and replace the q b's by a new letters b1, b2 ....., bq different from each other and also different from the rest. Obviously, these q new letters can be arranged among themselves in q! ways. Therefore, making such change in each of the x * p! permutations we will get x * p! * q! permutations in which p a's and q b's will be all different.

Therefore, replacing the r c's by r new letters c1,c2,......,cr and then per mutating them we will get x * p! * q! * r! permutations.

So, clearly the n letters are now all different. Therefore, x * p * q! * r! = the number of permutations of n different things taken all at a time = n! Theorem 3:

To find the number of permutations of n different things taken r at a time, when each thing may be repeated once, twice, ...., upto r times in any arrangement.

Supposing that we have r blank places and n different things where each thing may be taken once, twice, ..., upto r times. Then the required number of permutations is the same as the number of ways in which r blank places can be filled up by n different things

Then, the first place can be filled up in n ways, because any of the n may be put in it. When the first place is filled up, the second place can also be filled up in n ways, since the same letters may be used again. Thus the first two places can be filled up in n * n ways, i.e. n2 ways. The third place can also be filled up in n ways and, therefore, the first three places can be filled up in n2 * n = n3 ways.

Proceeding in this way and noticing that at any stage the index of n is always the same as the number of places filled up. Therefore, r blank places can be filled up by n different things in nr ways. Therefore, the required number of ways in which the r places can be filled is nr.

Example 1:

If 4 - x P 2 = 6, find the value of x?

Solution:

Since n P 2 = n (n - 1) hence, 4 - x P 2 = 6 gives, (4 - x) (4 - x - 1) = 6

Or, y(y - 1) = 6 [where 4 - x = y]

Or, y2 - y - 6 = 0

Or, (y - 3) (y + 2) = 06

Therefore, either, y - 3 = 0 i.e., y = 3 Or, y + 2 = 0          i.e., y = -2

Since y = 4 - x is positive, hence we take y = 3

Or, 4 - x = 3 [Since, y = 4 - x]           or, x = 1 Online Solution Different Formulas on Permutation Help:

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