# Nabeela Zubair's Question on Facebook (Counting Problem)

#### Sudharaka

##### Well-known member
MHB Math Helper
Nabeela Zubair on Facebook writes:

How I can solve this problem?

Students are choosing 2 colors to be used as school colors. There are 10 colors from which to choose. How many different ways are there to choose 2 different colors?

Total Colors 3 4 5 6 7 8 9 10
# of 2-color Comb. 3 6 10

#### Sudharaka

##### Well-known member
MHB Math Helper
Nabeela Zubair on Facebook writes:

How I can solve this problem?

Students are choosing 2 colors to be used as school colors. There are 10 colors from which to choose. How many different ways are there to choose 2 different colors?

Total Colors 3 4 5 6 7 8 9 10
# of 2-color Comb. 3 6 10
Hi Nabeela,

For the first color yo have 10 possibilities. After choosing the first color the second one sould be something different, so there are 9 possibilities for the second one. So by the multiplication principle there are $$10\times 9$$ total possibilities for choosing the two colors. But then there are repetitions involved here. That is we have counted each combination twice. If A and B are two colors we have counted A first, B second as well as B first, A second. Therefore the answer should be divided by two to get the number of combinations.

$\frac{10\times 9}{2}=45$

This idea can be generalized using the binomial coefficient.

$\binom {10}{2}=\frac{10!}{2!\,8!}=\frac{10\times 9}{2}=45$