95. In how many ways can 9 different items be distributed into three boxes so that one of them contains 3 items, another contains 2 items, and the third contains 4 items?
95. A n s w e r $. C_{9}^{3} \cdot C_{6}^{2}=9!:(3!\cdot 2!\cdot 4!)$. (Compare with problem 89. .