Question

We have 9 pens, of which 5 are green ink, 3 are red ink, and 1 is black. If we put the pens in a line, how many arrangements are possible

Answer 1

Answer:

504 arrangements are possible

Step-by-step explanation:

Arrangements of n elements:

The number of arrangements of n elements is given by:

$A_{n} = n!$

Arrangements of n elements, divided into groups:

The number of arrangements of n elements, divided into groups of $n_1, n_2,...,n_n$ elements is given by:

$A_{n}^{n_1,n_2,...,n_n} = \frac{n!}{n_1!n_2!...n_n!}$

In this case:

9 pens, into groups of 5, 3 and 1. So

$A_{9}^{5,3,1} = \frac{9!}{5!3!1!} = 504$

504 arrangements are possible