# hard combination/permutation problem. please help!

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• Oct 21st 2011, 04:15 PM
rsalort
hard combination/permutation problem. please help!
In a committee, there is only one chairman, 2 secretaries, and 3 members. If there are only 6 people, how many ways are there to assign them these positions? If there are 2 candidates for chairman, 4 candidates for secretary positions, and 5 candidates for the members position, how many ways are there to get such a committee?

please help
thank you
• Oct 21st 2011, 05:01 PM
rsalort
Re: hard combination/permutation problem. please help!
i did 6 pick 1 x 6 pick 2 x 6 pick 3= 3600

does anyone know if this is correct!?
any help is greatly appreciated
• Oct 21st 2011, 05:34 PM
Soroban
Re: hard combination/permutation problem. please help!
Hello, rsalort!

Quote:

In a committee, there is only one Chairman, 2 Secretaries, and 3 Members.
If there are only 6 people, how many ways are there to assign them these positions?

From the six people, choose one to be Chairman: . ${6\choose1} = 6$ choices.

From the remaining five people, choose two to be Secretaries: . ${5\choose2} = 10$ choices.

From the remaining three people. choose three to be Members: . ${3\choose3} = 1$ choice.

Therefore, there are: . $6\cdot10\cdot1 \:=\:60$ ways to form the committee.

Quote:

If there are 2 candidates for Chairman, 4 candidates for Secretary, and 5 candidates
for the Members positions, how many ways are there to get such a committee?

From the 2 Chairman candidates, choose one Chairman: . ${2\choose1} = 2$ choices.

From the 4 Secretary candidates, choose two Secretaries: . ${4\choose2} = 6$ choices.

From the 5 Member candidates, choose 3 Members: . ${5\choose3} = 10$ choices.

Therefore, there are: . $2\cdot6\cdot10 \:=\:120$ ways to form the committee.

• Oct 21st 2011, 05:39 PM
rsalort
Re: hard combination/permutation problem. please help!
thank you soroban, i really appreciate it
• Oct 21st 2011, 05:48 PM
agentmulder
Re: hard combination/permutation problem. please help!
for the 2nd part of the question, are there 11 people or still only 6?
• Oct 21st 2011, 06:02 PM
rsalort
Re: hard combination/permutation problem. please help!
Quote:

Originally Posted by agentmulder
for the 2nd part of the question, are there 11 people or still only 6?

there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc
• Oct 21st 2011, 06:16 PM
agentmulder
Re: hard combination/permutation problem. please help!
Quote:

Originally Posted by rsalort
there are 11 people for 6 positions, but not any person can take any position, only two can be chairman, only 3 can be secretary, etc

Oh, I see now, a person is not allowed to be a candidate for multiple positions in this problem.
(Wondering)