Question

Suppose we want to choose 5 colors, without replacement from 15 distinct colors, how many ways can this be done, if the order of the choices is :(a)not relevant and(b) is relevant?

Answer 1

Answer:

(a)3,003 ways.

(b)360,360 ways.

Explanation:

• The number of colors = 15

,

• The number of colors to be chosen = 5

(a)If the order of choices is not relevant we use the combination formula:

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

In this problem: n=15, x=5

$\begin{gathered} C_{15,5}=\frac{15!}{5!(15-5)!} \\ =\frac{15!}{5!\times10!} \\ =3003 \end{gathered}$

This can be done in 3,003 ways.

(b)If the order of choices is relevant we use the permutation formula:

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

In this problem: n=15, x=5

$\begin{gathered} P_{15.5}_{}=\frac{15!}{(15-5)!} \\ =\frac{15!}{10!} \\ =360,360 \end{gathered}$

This can be done in 360,360 ways.