Question

Suppose we want to choose 6 objects, without replacement, from 11 distinct objects.
(a) How many ways can this be done, if the order of the choices matters?
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(b) How many ways can this be done, if the order of the choices does not matter?

Answer 1

Answer:

a) 332640 ways

b) 462 ways

Step-by-step explanation:

Order:

If the order of the choices matters, we use the permutations formula. If they do not matter, we use the combinations formula.

Permutations formula:

The number of possible permutations of x elements from a set of n elements is given by the following formula:

$P_{(n,x)} = \frac{n!}{(n-x)!}$

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

6 objects, from a set of 11.

a) Order matters, so permutation.

$P_{(11,6)} = \frac{11!}{(11-6)!} = 332640$

b) Order does not matter, so combinations.

$C_{11,6} = \frac{11!}{6!(11-6)!} = 462$