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I first calculated the total amount of ways to have a committee of 6 members without the given restrictions and found that to be $(12*11*10*9*8*7/6!)$ = 924 Then I subtracted from 924 the total amount of ways that I could arrange of committee of 6 members that fails to meet the restrictions. I calculated the number of ways that a committee could be arranged in which the restrictions are not met to be equal to $4(9*8*7*6*5*4/6!)$ + 2 = 338 After subtracting 338 from 924 I obtained 586 as my final answer, however, the answer is D. 592 Would anyone like to take a crack at this? I have not taken Discrete Math, so if possible try to keep the solutions somewhat intuitive.

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Hints:

  • There are only $2$ ways to choose a committee of $6$ members such that one of the majors is not represented. In both these cases the $4$ classes are represented. So firstly disregard the condition that the majors must be represented and if that is solved then subtract $2$.

  • Disregarding majors $6=1+1+1+3$ and $6=1+1+2+2$ cover all possibilities. Handle these possibilities separately.

$592=\binom41\times\binom31\binom31\binom31\binom33+\binom42\times\binom31\binom31\binom32\binom32-2$

drhab
  • 151,093
  • How would I compute the total amount of committees composed of 6 members in which one of the classes is not represented? I got 336. Apparently that's what's messing up my answer. –  Dec 27 '14 at 12:51
  • not represented?.. Why do you want that amount? You must find the amount of committees of $6$ members in wich each of the classes is represented. That is by choosing $1$ class that is represented by $3$ members (and the other $3$ classes by $1$ member; the case $6=1+1+1+3$) or choosing $2$ classes that are both represented by $2$ members (and the other $2$ classes by by $1$ member; the case $6=1+1+2+2$). Choosing $1$ out of $4$ can be done on $\binom41=4$ ways and choosing $2$ out of $4$ in $\binom42=6$ ways. After this choosing you must choose the members out of these classes. – drhab Dec 27 '14 at 21:16
  • Oh, I understand what you're saying now. My initial approach was to find out the total number of 6-person committees that could be arranged in general and from that subtract the number of ways that I could arrange a 6-person committee that fails to meet the restrictions, leaving me with the number of 6-person committees that do meet the restrictions. After a while it did give me the correct answer but it was a pain. Your approach is much more straight-forward and efficient. Thank you! –  Dec 27 '14 at 22:06
  • You are very welcome. – drhab Dec 28 '14 at 07:57
  • You might want to add 'Disregarding majors...cover all possibilities' to second bullet. At first I thought those were the '2 ways' mentioned in the first bullet. That, coupled with the mouseover 'spoiler'(which at first I thought was an error) threw me for a tailspin at first. Nitpick is to replace #'s with their combos ${3\choose 1}/{3\choose 2},{3\choose 3}$ etc. Actually in this context non-trivial to state ${3\choose 1}={3\choose 2}=3$ and that one or the other is used. – miniparser Dec 28 '14 at 18:47
  • @underdog Is this edit an improvement? Main case stays of course that the OP understood me. – drhab Dec 29 '14 at 09:12
  • @drhab Yes. Thanks. – miniparser Dec 30 '14 at 01:05