96. In how many ways can 9 identical items be distributed into four boxes?
96. Let's add three more identical items, different from the given ones. We will then have 12 items of two types. From these, we can obtain $12!:(9!\cdot 3!)=$ $=C_{12}^{3}=220$ different permutations. In each such permutation, the items we added divide the given items into 4 parts (obviously, some of these parts may not contain any items). These 4 parts can be considered as the contents of four boxes. Answer. 220.