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You are in charge of creating a “USELESS” committee. The committee shall consist of one Umpire, three Statisticians, two Electricians, and one Lawyer (get it?). You personally know two Umpires, five Statisticians, three electricians, and four lawyers. Unfortunately, one of the Statisticians and one of the Electricians refuse to work together. How many USELESS committees can you create?

The answer given to me was 144.

However, my answer is 208

what i did was 208 = (combination of not having both) + (combination removed one statistician) + (combination removing one electrician)

=[(4c1)(2c2)(4c3)(2c1)]+[(4c1)(3c2)(4c3)(2c1)]+[(4c1)(2c2)(5c3)(2c1)]

=32+96+80 =208

Thank you

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You're double-counting some possibilities. For example every committee without either of the troublesome members is counted once in each of your three terms.

It would be easier to do this by subtraction than by addition -- first count the total number of legal committees, then subtract those that have both troublemakers on them:

$$ \binom21_U \binom 53_S \binom32_E \binom 41_L - \binom21_U \binom42_S \binom 21_E \binom 41_L $$

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  • $\begingroup$ Ok last question why would it be (4c2) for the statistician and (2c1) for the electrician, when you need 3 members for the statistician and 2 members for the electrician, which respectively is (4c3) (2c2)? $\endgroup$ Oct 13 '13 at 21:15
  • $\begingroup$ @John: The second term counts the committees where the troublesome statistician is on the committee, so the only thing to choose is which of the four other ones are in the two remaining seats. Similarly for electricians. $\endgroup$ Oct 13 '13 at 21:22

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