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A Mathematics department consists of 5 female and 5 male teachers.

How many committees of 3 teachers can be chosen which contain at least one female and at least one male? a) 100 b) 120 c) 200 d) 2500

So I found several ways of doing this question Method 1: 10C3 - 5C3 - 5C3 = 100 (total number of combinations - number of combinations with no male - number of combination with no female)

Method 2: 2x(5C1 x 5C2) = 100 [number of combinations with 1 male and 2 female multiplied by 2 (as it can be 1 female and 1 male)]

These two both work to give the correct answer but the following method was my initial attempt.

My Method: 5C1 x 5C1 x 8C1 = 200 [The number of combinations of choosing 1 from male x number of combinations of choosing 1 from female x number of ways you can choose 1 from the remaining 8 people]

Can you please explain why my method does not work?

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    $\begingroup$ The problem is that the third method counts things twice: for example, male 1 could be either the required male or the optional male. $\endgroup$ – rogerl Aug 29 '14 at 1:35
  • $\begingroup$ I still don't quite get it. By saying the last one is 8C1 not 10C1, aren't you eliminating the fact that the last option is exclusive of the 2 already chosen? $\endgroup$ – Charlie Aug 29 '14 at 1:45
  • $\begingroup$ With your method, you could choose John as the required male and Jim as the optional third person, or Jim as the required male and John as the optional third person. But those are the same committee. Thus you have overcounted. $\endgroup$ – rogerl Aug 29 '14 at 1:47
  • $\begingroup$ Ahhh, It took me a while to register but I get it now. Thank you very much :) $\endgroup$ – Charlie Aug 29 '14 at 2:00
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    $\begingroup$ In this case, the third method will work, as long as one is aware that every committee is counted exactly twice, so we need to divide by $2$ at the end. There are many instances where deliberate multiple-counting, followed by adjustment, is an effective strategy. $\endgroup$ – André Nicolas Aug 29 '14 at 5:13
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You are double counting each committee. If the men are ABCDE and the women FGHIJ, the committee CHJ gets counted once by choosing C from the men, H from the women, and J from the rest and again by choosing C from the men, J from the women, and H from the rest.

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