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Twelve doctors are to be selected by hospital staff to sit on a six-person committee. Within that committee, an additional subcommittee of three doctors will be formed. How many unique sub-committees are possible?...... Please let me know

atir
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    Welcome to Math.SE! Please consider updating your question with some information about what you have tried or where you are getting stuck. You will find people are much more willing to help if you do! – Nick Peterson Nov 05 '13 at 19:05
  • i think so basically in the end the question implies that "no of unique committee of 3 guys from 12 " which is equal to 12C3 ...correct me if am wrong... thanks – atir Nov 05 '13 at 19:16

2 Answers2

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Since we want the number of unique subcommittees we know that there are ${12\choose 3}=220$ unique subcommittees. The reason we know this is because in total there are $12$ doctors and a subcommittee has $3$ doctors in it. We want all of the unordered selections of $3$ of the $12$ doctors.

1233dfv
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  • You're welcome! :) It was an interesting problem. – 1233dfv Nov 05 '13 at 20:21
  • yeah the question misguides you in thinking that we should first find no of committees i.e 12C3 , multiply by number of unique 3 guys sub-committee 6c3 ...but actually its not that case... do you know any such problem where this sort of nesting of committees it would have depended on each other? – atir Nov 05 '13 at 20:27
  • If they asked for how many different committees with subcommittees of the specified sizes we would know that there are ${12\choose 6}\cdot{6\choose 3} committees with subcommittees. In combinatorics there are similar identities where a committee argument can be used to prove the validity of an identity. For example the identity $k{n\choose k}=n{n-1\choose k-1}$ and the identity ${r\choose m}\cdot{m\choose k}={r\choose k}\cdot{r-k\choose m-k}$. – 1233dfv Nov 05 '13 at 20:38
  • If they asked for how many different committees with subcommittees of the specified sizes we would know that there are ${12\choose 6}{6\choose 3}$ committees with subcommittees. In combinatorics there are similar identities where a committee argument can be used to prove the validity of an identity. For example the identity $k{n\choose k}=n{n-1\choose k-1}$ and the identity ${r\choose m}{m\choose k}={r\choose k}{r-k\choose m-k}$. – 1233dfv Nov 05 '13 at 20:45
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We use the product rule, first there are ${{12}\choose{6}}$ ways of forming six-person committees. From each six-person committee, ${{6}\choose{3}}$ sub-committees can be formed.

Hence, by the product rule, the total number of committees is ${{12}\choose{6}}{{6}\choose{3}}$!

Nedellyzer
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  • but this will lead to duplication right!? coz the same 3 guys sub committee can be formed for different combinations of 6 guys committee...so there will be duplication....leme know what are your thoughts!? – atir Nov 05 '13 at 19:25
  • That is correct. Consider the two sets with $6$ doctors ${d_1,d_2,d_3,d_4,d_5,d_6}$ and ${d_1,d_2,d_3,d_4,d_5,d_7}$. There are subsets of $3$ doctors that are the same for both of these sets. Using the Multiplication Principle like this leads to over counting. We need the subsets of size $3$ to be unique. – 1233dfv Nov 05 '13 at 20:55