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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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To be picky, what you wrote is an answer, not a solution. Are the objects distinct or not? I assume the people are to be considered distinct! If the objects are distinct, your answer is off by a factor of $4$. Perhaps you did not take into account the fact that the person who will receive $1$ object can be chosen in $\binom{4}{1}$ ways. –  André Nicolas Jan 10 '12 at 19:16
    
yes they are different. thank you. –  Oluwagbenga Jan 10 '12 at 19:19
    
If you can produce an argument that the answer (after calculation) is $2520$, your argument is likely to be correct. –  André Nicolas Jan 10 '12 at 19:22
    
@Andre, can you please show me the calculation? –  Oluwagbenga Jan 10 '12 at 19:39

1 Answer 1

up vote 2 down vote accepted

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. –  Oluwagbenga Jan 10 '12 at 20:27

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