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

(a)3,003 ways.

(b)360,360 ways.

Step-by-step 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)=(n!)/(x!(n-x)!)`

In this problem: n=15, x=5

`\begin{gathered} C_(15,5)=(15!)/(5!(15-5)!) \\ =(15!)/(5!*10!) \\ =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)=(n!)/((n-x)!)`

In this problem: n=15, x=5

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

This can be done in 360,360 ways.