see ans of joy is incorrect because he has considered the order in which the toys are given to the child as different cases

for eg: if a child has two toys after final distribution say toy 1 & toy 2 it should be one case but in joy's method it is two cases,

1st- toy1 from first 8 lot and toy2 from the other remaining 2 toys

2nd- toy 2 from the first 8 lot and toy 1 from the remaining two.

hence two eliminate this problem of order recurrence we shall take two cases

first case let the three toys be given to a single person

no.of ways 10C2 * 8C1 * 10C7 * 7!           

case 2 remaining two toys are given to two different persons(i.e 6 persons get one each and two get two toys each)

no.of ways 10C2 * 8C2 * 10C6 * 6! * 4C2

total ans  = case 1 + case2