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Consider 7 different objects and 4 persons. How many distributions of the objects are there if we require that 3 persons receive 2 objects and 1 person receives 1 object?

I solved as follows $\frac{7!}{2!2!2!1!}$. Am I correct?

Thanks

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    @Andre, can you please show me the calculation?2012-01-10

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The person who will receive only one gift can be chosen in $\binom{4}{1}$ ways. For each such choice, the actual gift she will receive can be chosen in $\binom{7}{1}$ ways. So there are $\binom{4}{1}\binom{7}{1}$ ways to select the person who will receive $1$ gift only, and the gift she will receive.

Once we have done this, we have $6$ gifts left, and we need to give $2$ of these to each of $3$ people.

Call the $3$ lucky people A, B, and C. The gifts for A can be chosen in $\binom{6}{2}$ ways. For each such way, there are $\binom{4}{2}$ ways select the gifts for B. And once we have done that, what C gets is determined. If you like, there are then $\binom{2}{2}$ ways to decide what C gets. It follows that the total number of ways of doing the job is $\binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}.$ Finally, calculate. We get $2520$.

Comment: You may have been using multinomial coefficients, or some related formula. First choose who will get only one gift. This can be done in $\binom{4}{1}$ ways. Suppose that for example the people are called A, B, C, D, and D is to get only one gift. So we want to distribute $7$ objects in the pattern $2$-$2$-$2$-$1$. By a remembered formula, the number of ways to do this is $\binom{7}{2,2,2,1}\qquad\text{that is,}\qquad \frac{7!}{2!2!2!1!}.$ So our total count is $\binom{4}{1}\frac{7!}{2!2!2!1!}.$ From the answer you arrived at, I assume that your reasoning was along the lines of this comment, and only the factor $\binom{4}{1}$ was missing.

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    thank you very much for your tome and help. I now understand fully. I will try to practice more with this reasoning.2012-01-10