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A committee of size 4 is to be randomly formed from 5 professors, 7 associate professors and 11 assistant professors. What is the probability that a committee consists of a person from each rank?

Attempt: So i said there would be $ 5 . 7 . 11 . 20 $ possibilities. Dividing this by the sample space$ ($ gives the incorrect answer. Why?

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What is conceptual in there? – Did Dec 2 '12 at 11:23
up vote 2 down vote accepted

There are a couple of problems with your approach. First, $23\cdot22\cdot21\cdot20$ is not the number of ways to choose a four-person committee; it’s the number of ways to choose a four-person committee in a particular order. Since each committee can be arranged in $4!=24$ different orders, you’ve counted each committee $24$ times.

The calculation $5\cdot7\cdot11\cdot20$ takes the order of the choices into account, but not in the same way: here you’re definitely choosing a professor first, then an associate professor, then an assistant professor, then a fourth committee member. This counts each possible committee exactly twice. To see this, consider a committee consisting of professors A and B, associate professor C, and assistant professor D: the calculation $5\cdot7\cdot11\cdot20$ counts it once for the order ACDB and once again for the order BCDA.

Thus, the fraction $$\frac{5\cdot7\cdot11\cdot20}{23\cdot22\cdot21\cdot20}$$ counts each acceptable committee twice in the numerator and each possible committee $24$ times in the denominator. This clearly doesn’t give the ratio of acceptable to possible committees, which is the probability of picking an acceptable committee. To get the right result, divide the numerator by $2$ and the denominator by $24$ to correct for the overcounting.

Here’s possibly a more straightforward way to arrive at the correct result.

There are three kinds of acceptable committees: those that have two professors, those that have two associate professors, and those that have two assistant professors.

  • There are $\binom52\cdot7\cdot11=770$ committees with two professors, one associate professor, and one assistant professor.
  • There are $\binom72\cdot5\cdot11=1155$ committees with two associate professors, one professor, and one assistant professor.
  • There are $\binom{11}2\cdot5\cdot7=1925$ committees with two assistant professors, one professor, and one associate professor.

Since there are altogether $\binom{23}4=8855$ four-person committees, the probability of getting an acceptable one at random is


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Thanks a lot for this reply. When I said, why do I not need to divide by 4! here since we don't care about the ordering of the choosing? – CAF Dec 2 '12 at 11:20
@CAF: You’re welcome. The difference is in the amount of overcounting: in the denominator you counted each possible $4$-person committee $4!$ times, but in the numerator you counted each acceptable committee only twice. – Brian M. Scott Dec 2 '12 at 11:23

It is simpler to calculate it using permutations instead of combinations. We want the probability of ordered subsets that have one member of each of the three types plus one of the remaining 20. The numerator of the probability is simply the product of the possibilities of one of each type, times one of the remaining 20, times the possible arrangements which equals 5x7x11x20x12. The 12 is the number of possible arrangements of 3 distinct items and one duplicate item or 4! / 2!. The denominator is simply all arrangements of 4 of the 23 or 23x22x21x20. The result after cancellation is 10/23.

Permutations are often considerably simpler to calculate than combinations.

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