# Question on permutation and combination

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?

• Could you explain the sense of "not determined"?
– whuber
Commented May 12, 2023 at 20:18
• @whuber Exactly the scenario given in the highlighted problem statement Commented May 12, 2023 at 20:33
• I'm afraid that doesn't explain anything. Something is the matter with your interpretation of the question, so please elaborate on that interpretation.
– whuber
Commented May 12, 2023 at 20:48
• 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? Commented May 12, 2023 at 20:54
• 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.)
– whuber
Commented May 12, 2023 at 20:55

• +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!) $\binom{4}{1;2;1}=4!/(1!2!1!).$