PERMUT

Returns the number of ordered permutations for a given number of objects.

Syntax:

PERMUT(n, k)

where n and k are integers.

PERMUT returns the number of ordered ways that k objects can be chosen from a set of n objects, where an object can only be chosen once. For example with a set of 3 objects A, B, C, we can choose 2 as follows: AB, AC, BA, BC, CA, CB.

PERMUT calculates:

$_{n} P_{k} = \frac{n !}{( n - k )!}$

Example:

PERMUT(3, 2)

returns 6, as in the example above.

Application:

Let's imagine a running race with 10 participants. We want to find out how many different ways the top 3 spots (1st, 2nd, and 3rd place) can be filled.

In this scenario:

n (the total number of items) is 10 (the total number of runners).
k (the number of items being chosen in a specific order) is 3 (for 1st, 2nd, and 3rd place).

The PERMUT function calculates this as $P (n, k) = \frac{n !}{( n - k )!}$ .

Using our example, the calculation is:

$P (10, 3) = \frac{10 !}{( 10 - 3 )!} = \frac{10 !}{7 !} = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 10 \times 9 \times 8 = 720$

There are 720 different possible ways for the 1st, 2nd, and 3rd place spots to be awarded to the 10 runners.

Here is a table to illustrate the concept:

	Rank	Number of Choices (from the remaining runners)
	A	B
1	1st Place	10
2	2nd Place	9
3	3rd Place	8

This shows that for the first-place position, any of the 10 runners can win. Once the first-place winner is determined, there are only 9 runners left who can win second place. Finally, there are 8 runners left to win third place. The total number of permutations is found by multiplying these choices together: 10×9×8=720.