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I have the following question:

A committee of 4 person is to be chosen from 8 person, including Mr.Smith and his wife. Mr.Smith will not join the committee without his wife, but his wife will join without him. Calculate the number of ways in which the committee of 4 person can be formed.

My Logic is that this must be a combination question since John,Jane,Mr.Smith and Mrs.Smith is the same in any order. Given that I find the combination of this by $8*7*6*5= 1680$, I then divide it by $4!$ to get $70$ total combinations.

This is where it gets a little tricky for me , Mr.Smith wont join if his wife isnt on it so I decided to make him be counted as one person found the combination from that so $7*6*5*4 = 840$ Divide that by $4!$ and then subtract it from the $70$ which equal $35$ ways to rearrange the committee. Does this make sense?

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Split it into disjoint cases, and then add them up:

  • The number of committees $\color\red{\text{with}}$ Mr. Smith and $\color\red{\text{with}}$ Mrs. Smith is $\binom{8-2}{4-\color\red2}=15$
  • The number of committees without Mr. Smith and $\color\red{\text{with}}$ Mrs. Smith is $\binom{8-2}{4-\color\red1}=20$
  • The number of committees without Mr. Smith and without Mrs. Smith is $\binom{8-2}{4-\color\red0}=15$
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  • $\begingroup$ BTW, the probability tag on your question seems redundant... $\endgroup$
    – barak manos
    Commented Oct 31, 2016 at 8:49
  • $\begingroup$ And BTW, last time I heard, Mr. and Mrs. Smith are no longer together... $\endgroup$
    – barak manos
    Commented Oct 31, 2016 at 8:50
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    $\begingroup$ PS: Alternatively, there are $\binom 8 4$ ways to select a committee of which $\binom {8-2}{4-1}$ are the excluded case (selecting the Mr. without the Mrs.). Subtraction gives the same answer as summing the above. $\endgroup$
    – Graham Kemp
    Commented Oct 31, 2016 at 8:59
  • $\begingroup$ @GrahamKemp: Yes, that's even a simpler (shorter by one bullet) solution in this case. $\endgroup$
    – barak manos
    Commented Oct 31, 2016 at 14:01

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