Distributing objects question 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
 A: 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. 
