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A certain board has 12 persons: a president, two vice-presidents, eight managers, and state inspector. A committee of 6 people needs to be elected. If two vice-presidents must be on the same committee and the inspector and the president cannot be on the same one, how many different committees can be formed?
тАв 28
тАв 160
тАв 210
тАв 170
тАв 182

Can anybody provide some explanations?

Last edited by bb on 30 Jul 2003, 18:19, edited 1 time in total.

A certain board has 12 persons: a president, two vice-presidents, eight managers, and state inspector. A commission of 6 people needs to be elected. If two vice-presidents have to be on the same team and the inspector and the president cannot be on the same team, how many different committees can be formed? тАв 28 тАв 160 тАв 210 тАв 170 тАв 182

Can anybody provide some explanations?

This problem is confusing. You use 3 different words for subgroups of the commission: commission, committee, and team. Are they all the same thing?
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

A certain board has 12 persons: a president, two vice-presidents, eight managers, and state inspector. A commission of 6 people needs to be elected. If two vice-presidents have to be on the same team and the inspector and the president cannot be on the same team, how many different committees can be formed? тАв 28 тАв 160 тАв 210 тАв 170 тАв 182

Can anybody provide some explanations?

Sorry, I thought I revised the problem - it is not mine - one from long time ago and I have a point of confusion about it. All those things mean the same thing. I edited the original to be more clear.

Let two seats on the committee we reserve for vice presidents.
Then we got 10C4=210 combinations for the other four seats.
Now we can calculate the number of committees with the president and the inspector 8C2=28.
The answer is 210-28=182.

Let two seats on the committee we reserve for vice presidents. Then we got 10C4=210 combinations for the other four seats. Now we can calculate the number of committees with the president and the inspector 8C2=28. The answer is 210-28=182.

That is probably the best method.

An alternate method:

We can have either the president (no inspector), the inspector (no president), or neither of them on the committee so we can count number of ways for each situation, then add them up.

Let's assume the president is on:
Then we need to choose 3 managers out of 8 to fill the seats.
8C3 = 56.
If the inspector is on the committee, the calculation is the same, 8C3 = 56.
Now if both are not on, we need to pick 4 of 8 managers, or 8C4 = 70.

56 + 56 + 70 = 182.

QED.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: This problem is unclear - let me explain why [#permalink]

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31 Jul 2003, 13:19

htown wrote:

I know it says that the two VPs must be on the same team, but it never says that all teams have to have the VPs.

I would assume that a committee of 6 managers is possible given the wording of the question.

Can someone please tell me where I am going wrong with this?

Thanks

Htown

YOu have a good point and I thought about this because, as I mentioned, the question is poorly worded.

As far as I can tell, there is only one committee, so I made an assumption that they would both be on that committee. (i.e., if no VPs, then the VP would not be on the same team because they are not on any team). However, that is just my interpretation of the question.
_________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993