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# Ben needs to form a committee of 3 from a group of 8

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Ben needs to form a committee of 3 from a group of 8 [#permalink]

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23 Apr 2005, 15:10
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers cannot serve together, how many different committees can Ben form?
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23 Apr 2005, 15:21
If two engineers cannot serve together we have to split them.

X = N(problem engineers not picked at all) +
N(first problem engineer picked) +
N(second problem engineer picked) =
3C6 + 2 * 2C6 =
20 + 2 * 15 = 50
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23 Apr 2005, 18:01
Can you explain how you get 6C2?
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23 Apr 2005, 21:24
If one position is filled with problem engineer we are left with two to pick from 6 nonproblem engineers.
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24 Apr 2005, 08:07
total - unfavorable
8c3-6c1=56-6=50
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25 Apr 2005, 09:38
8x7x6 = 336 (total arrangements)
3x2x1 = 6 (each group working individually)

336
---- = 56
6
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25 Apr 2005, 10:41
rc1979 wrote:
Ben needs to form a committee of 3 from a group of 8 engineers. If two of the engineers cannot serve together, how many different committees can Ben form?

Assume A and B are these two gentlemen in question.

Total ways of selecting engineers = C(8,3) = 56.
This includes
1. A included.
2. B included.
3. A and B included.
4. Nither A or B included.

Of course we need to discard the case of A and B included which is

C(6, 1) = 6.

Therefore, remaining cases, when these two gentlemen don't serve together = 56 - 6 = 50.
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25 Apr 2005, 10:59
kapslock,
can you explain how you get 6c1?
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25 Apr 2005, 14:46
rc1979 wrote:
kapslock,
can you explain how you get 6c1?

RC,

This is the case of number of selections when A and B are both included.

If A and B are included, you need to select only one more guy out of 6 remaining guys to complete the selection of 3 (because 2 out of the total 8 are A and B themselves).

Thus we have a case of selection of 1 from 6, which is C(6,1) = 6.

Hope that helps.
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25 Apr 2005, 14:46
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