Welcome to our community

Be a part of something great, join today!

Nabeela Zubair's Question on Facebook (Counting Problem)

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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
Feb 5, 2012
1,621
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.

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