The toys are differnt here but the boxees are identical.
since no box can be empty we can have two situations for this.
A)2,2,1 B)3,1,1 considering case A(2,2,1) ways of selecting 2 toys from 5 = 5C2=5.4/1.2=10 then to select 2 from the rest three = 3C2 = 3.2/1.2=3
total number of ways to puting this into boxes in this case = 5C2 * 3C2 * 1C1 = 30
similarly,considering case B(3,1,1)
total number of ways to puting this into boxes in this case = 5C3 * 1C1 * 1C1 = 10
so total = 30 + 10 = 40 ---> wrong
this seems incorrect. what would be the proper solution.