The number of ways in which four different toys and five indistinguishable marbles can be distributed between 3 boys, if each boy receives at least one toy and at least one marble, is

Options

A. 42
B. 100
C. 150
D. 216

Answer

To solve the problem of distributing four different toys and five indistinguishable marbles among three boys, while ensuring that each boy receives at least one toy and at least one marble, we can break the problem down into two parts: distributing the marbles and distributing the toys.

### Step 1: Distributing the Marbles
We have 5 indistinguishable marbles and 3 boys. Since each boy must receive at least one marble, we can start by giving one marble to each boy. This uses up 3 marbles, leaving us with 2 marbles to distribute freely among the 3 boys.

To find the number of ways to distribute these 2 indistinguishable marbles among 3 boys, we can use the "stars and bars" combinatorial method. The formula for distributing \( n \) indistinguishable objects into \( r \) distinct boxes is given by:

\[ \text{Number of ways} = \binom{n + r - 1}{r - 1} \]

In our case, \( n = 2 \) (the remaining marbles) and \( r = 3 \) (the boys). Thus, we have:

\[ \text{Number of ways to distribute marbles} = \binom{2 + 3 - 1}{3 - 1} = \binom{4}{2} = 6 \]

### Step 2: Distributing the Toys
Next, we have 4 different toys to distribute among the 3 boys, ensuring that each boy gets at least one toy. The only valid distributions of toys that satisfy this condition are:

1. One boy gets 2 toys, and the other two boys get 1 toy each.

Let's denote the boys as A, B, and C. We can choose which boy will receive 2 toys in \( 3 \) ways (A, B, or C).

After selecting the boy who will receive 2 toys, we can choose 2 toys from the 4 available toys. The number of ways to choose 2 toys from 4 is given by:

\[ \binom{4}{2} = 6 \]

The remaining 2 toys will automatically go to the other two boys, with each receiving 1 toy. The number of ways to assign these 2 toys to the 2 boys is \( 2! = 2 \).

Thus, the total number of ways to distribute the toys is:

\[ \text{Total ways to distribute toys} = 3 \times \binom{4}{2} \times 2! = 3 \times 6 \times 2 = 36 \]

### Step 3: Combining the Results
Since the distribution of marbles and toys are independent events, we can multiply the number of ways to distribute the marbles by the number of ways to distribute the toys:

\[ \text{Total ways} = \text{Ways to distribute marbles} \times \text{Ways to distribute toys} = 6 \times 36 = 216 \]

Thus, the total number of ways to distribute the four different toys and five indistinguishable marbles among three boys, ensuring each boy receives at least one toy and one marble, is **216**.

### Final Answer
**216**