5. There are $n$ different items and 9 numbered boxes. How many ways are there to distribute these items into the boxes so that the first box contains as many items as the sum of the items in the boxes from the second to the fifth?
5. Solution. Let there be exactly $k$ items in the first box. Then the number of ways to distribute the items is $C_{n}^{k}$. $C_{n-k}^{k} \cdot 4^{k} \cdot 4^{n-2 k}=C_{n}^{k} \cdot C_{n-k}^{k} \cdot 4^{n-k}$. Now we need to sum such numbers over all permissible $k$. This sum will be the free coefficient in the expression $\left(x+\frac{4}{x}+4\right)^{n}$. The latter is equal to $\frac{(x+2)^{2 n}}{x^{n}}$. The free coefficient here, obviously, is $2^{n} C_{2 n}^{n}$.