Question

Find both the number of combinations and the number of permutations for the given number of objects.

Answer 1

Answer:

First, if we have a set of K elements, such that are ordered as:

{x₁, x₂, ...}

The total number of permutations for the K elements can be found in the next way.

For the first element in the set, we have K options.

For the second element in the set, we have (K - 1) options (because we already choose one)

For the third element we have (K - 2) options, and so on.

The total number of permutation is equal to the product between the numbers of options for each position's element, then the number of permutations for K elements is:

permutations = K*(K - 1)*(K - 2)*....*2*1 = K!

Now suppose that we have a set of N elements, and we want to make groups of K elements.

The total number of different combinations of K elements is given by the equation:

$C(N, K) = \frac{N!}{(N - K)!*K!}$

In this case we have 15 objects (then  N = 15) and we take 7 at the time (Then K = 7)

Where we need to take in account the number of combinations and also the permutations for each combination.

Then the total number of different sets is:

C(15*7)*7!

First, the total number of combinations will be:

$C(15,7) = \frac{15!}{(15 - 7)!7!} = \frac{15!}{8!*7!} = \frac{15*14*13*12*11*10*9}{7*6*5*4*3*2*1} = 6,435$

So we have 6,436 combinations, and each one of these combinations has 7! permutations.

permutations = 7! = 7*6*5*4*3*2*1 = 5,040

if we combine these we get:

Combinations*Permutations = 6,435*5,040 = 32,432,400