Number of ways in which four different toys and five indistinguishable

Solution:

Let’s begin with the number of ways to distribute the toys. Since there are three children and four toys, and since each child will receive at least one toy, then one child must get two toys and the remaining two children must get one toy. If Amar is the child to get two toys, there are 4C2 = 6 ways to choose the toys that he will receive. Akbar will get one of the two remaining toys, so there are 2 ways to choose the toy that he will receive. Anthony will receive the remaining toy. Thus, there are 6 x 2 = 12 ways in which Amar gets two toys and the remaining children get one toy each.

The number of ways where Amar gets two toys and the number of ways where Anthony gets two toys are also 12; both can be calculated with a similar reasoning as above. Thus, there are 12 + 12 + 12 = 36 ways to distribute the toys.

Now, let’s calculate the number of ways to distribute the marbles. Since the marbles are identical and since each child will receive at least one marble, let’s give each child one marble and determine the number of ways we can distribute the remaining two marbles among the three children such that each child gets 0, 1 or 2 marbles. Notice that there are two scenarios: 1) one child gets two marbles and the remaining children don’t get any marbles, or 2) two children get one marble each and the remaining child doesn’t get any marbles. In the former case, there are three choices for the child to get two marbles, and in the latter case, there are three choices for the child to get zero marbles. Thus, there are 3 + 3 = 6 ways to distribute the two marbles among the children after each child gets one marble.

Since there are 36 ways to distribute the toys and since there are 6 ways to distribute the marbles, there are 36 x 6 = 216 ways to distribute the toys and the marbles.

Answer: D