Question

Suppose we want to choose 4 objects, without replacement, from 16 distinct objects (a) How many ways can this be done, if the order of the choices is not relevant? (b) How many ways can this be done, if the order of the choices is relevant?

Answer 1

Answer:

a) 1820 ways

b) 43680 ways

Step-by-step explanation:

When the order of the choices is relevant we use the permutation formula:

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

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

When the order of choices is not relevant we use the combination 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 problem, we have that:

$x = 4, n = 16$

(a) How many ways can this be done, if the order of the choices is not relevant?

$C_{16,4} = \frac{16!}{4!(12)!} = 1820$

(b) How many ways can this be done, if the order of the choices is relevant?

$P_{16,4} = \frac{16!}{(12)!} = 43680$