I am having a trouble understanding how to answer the following question and what method to use. The final answer is $415800$

Suppose there are 4 committees A, B, C, and D. 11 candidates are randomly assigned to these 4 committees. Each candidate can only be assigned to 1 committee. In how many ways can we randomly assign the 11 candidates to these four committees such that one committee consists of 1 member, one committee consists of 4 members, another committee consists of 4 members, and another committee consists of 2 members?

Now if I know how many candidates are needed for each of $A,B,C,D$ committees (e.g., 1, 4,4,2 candidate for $A,B,C$,and $D$, respectively, the probability will be

$$11 \cdot {10\choose 4} \cdot {6\choose 4} \cdot {2\choose 2} = 34650$$

But how do I deal with the fact that they are not determined?

  • $\begingroup$ Could you explain the sense of "not determined"? $\endgroup$
    – whuber
    Commented May 12, 2023 at 20:18
  • $\begingroup$ @whuber Exactly the scenario given in the highlighted problem statement $\endgroup$
    – Rudinberry
    Commented May 12, 2023 at 20:33
  • $\begingroup$ I'm afraid that doesn't explain anything. Something is the matter with your interpretation of the question, so please elaborate on that interpretation. $\endgroup$
    – whuber
    Commented May 12, 2023 at 20:48
  • $\begingroup$ Does this phrase better "How do I incorporate the assumption that randomly 2 committees must have 4 candidates, 1 committee has 2 candidates and 1 committee has 1 candidate? $\endgroup$
    – Rudinberry
    Commented May 12, 2023 at 20:54
  • $\begingroup$ No, because the answer you derive has already done that. Indeed, 415800 is six times 34650. It is not apparent where that "final answer" comes from and why it multiplies your result by six. What justifies it? (BTW, your answer is not a "probability:" probabilities, by definition, are between 0 and 1.) $\endgroup$
    – whuber
    Commented May 12, 2023 at 20:55

1 Answer 1


You need to multiply your answer by the number of ways in which you can make the four committees, i.e. the number of ways in which 1,4,4,2 can be clubbed with A, B, C, D. This can be done in 12 ways (= the number of 4 digit numbers that you can make with 1,4,4,2 = 4!/2!)

So the final answer should be = 34650 * 12 = 415800

  • $\begingroup$ That's already accommodated in the calculation of 34650. Thus, multiplication by 12 is superfluous and doesn't give the correct answer. $\endgroup$
    – whuber
    Commented May 18, 2023 at 22:22
  • $\begingroup$ Calculation of 34650 only accounts for 1 permutation and not all 12 combinations. $\endgroup$ Commented May 19, 2023 at 1:47
  • $\begingroup$ +1 I believe your interpretation reads the question not as "committee $A$ has $1$ person, committee $B$ has $4$ people," and so on, as I have done. Instead (quite unusually) the committee sizes have not been fixed beforehand E.g., committee $A$ might have $1,$ $2,$ or even $4$ members; and likewise for the others. The possibilities can be counted in two stages: assign $11$ people to unlabeled committees of sizes $1,4,4,2;$ and then label the committees. The factor $12$ comes from the multinomial coefficient (not the binomial coefficient!) $\binom{4}{1;2;1}=4!/(1!2!1!).$ $\endgroup$
    – whuber
    Commented May 19, 2023 at 14:13

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