The number of ways in which 10 different toys can be distributed to 8 children so that each child gets at least one toy is

Two cases. i) One child gets 3 toys and 7 children get one toy each. $\frac{10!}{3!7!(1!{)}^7}\times 8!$

ii) Two children get 2 toys each and 6 children get toy each $\frac{10!}{2!(2!{)}^2\times 6!(1!{)}^6}\times 8!\therefore \text{ Ans }=960\times 8!$