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The UOT Society has an Red Committee (RC) consisting of five members and a Blue Committee (BC) consisting of six members.

• Assume there are 6-11 members in the Society, represented by $n$. Also, assume that a member can be both on the RC and on the BC. What is the total number of ways in which these two committees can be chosen?

I believe I'm supposed to look at each value of $n$, but I'm not really sure. so..

if $n = 6$: $${6\choose 5} \times {6\choose 6}$$ if $n = 7$: $${7\choose 5} \times {7\choose 6}$$ if $n = 8$: $${8\choose 5} \times {8\choose 6}$$ etc. I'm not sure what to do after this. Do I just add all the results? Any help appreciated.

Peter Woolfitt
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saeid
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1 Answers1

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Yes, that's correct. The wording is a little unclear, but to get the total number, you would just add them as you suggest, and now it's just a calculation problem.

Peter Woolfitt
  • 21,114