7 distinct toys are distributed among 4 boys in such a way that any three get each 2 toys and forth gets 3 toys. Find No of ways.

Answer

To solve the problem of distributing 7 distinct toys among 4 boys such that any 3 boys receive 2 toys each and the fourth boy receives 3 toys, we can follow these steps:

### Step-by-Step Solution:

1. **Choose 3 Boys from 4**:
First, we need to select 3 boys out of the 4 to receive 2 toys each. The number of ways to choose 3 boys from 4 is given by the combination formula \( \binom{n}{r} \), where \( n \) is the total number of boys and \( r \) is the number of boys to choose.

\[ \text{Number of ways to choose 3 boys} = \binom{4}{3} = 4 \]

**Hint**: Remember that the combination formula is used when the order of selection does not matter.

2. **Distribute Toys to the Chosen Boys**:
After selecting 3 boys, we need to distribute 6 toys (2 toys each) to them. The number of ways to select 2 toys for the first boy from the 7 toys is \( \binom{7}{2} \). After giving 2 toys to the first boy, we have 5 toys left. We then select 2 toys for the second boy from these 5 toys, which is \( \binom{5}{2} \). Finally, the last boy will receive the remaining 2 toys.

\[ \text{Number of ways to distribute toys} = \binom{7}{2} \times \binom{5}{2} \times \binom{3}{2} \]

**Hint**: Use the combination formula to find the number of ways to choose toys for each boy sequentially.

3. **Calculate the Remaining Toys for the Fourth Boy**:
The fourth boy will automatically receive the remaining 3 toys. Since we have already accounted for the distribution to the first three boys, there is no need for further calculations here.

4. **Calculate the Total Ways to Distribute the Toys**:
Now, we need to multiply the number of ways to choose the boys by the number of ways to distribute the toys.

\[ \text{Total ways} = \binom{4}{3} \times \left( \binom{7}{2} \times \binom{5}{2} \times \binom{3}{2} \right) \]

5. **Factorial Arrangement of Boys**:
Since the boys are distinct, we can arrange them in \( 4! \) ways. However, since we have given 2 toys to each of the 3 boys, we need to divide by \( 2! \) for each of these boys (3 boys) to account for the indistinguishable arrangements of the toys they received. The fourth boy's toys can be arranged in \( 3! \) ways.

\[ \text{Final Count} = \text{Total ways} \times \frac{4!}{(2!)^3 \times 3!} \]

6. **Final Calculation**:
Now we can compute the values:

\[ \binom{7}{2} = 21, \quad \binom{5}{2} = 10, \quad \binom{3}{2} = 3 \]

Therefore,

\[ \text{Total ways} = 4 \times (21 \times 10 \times 3) = 4 \times 630 = 2520 \]

And the arrangement factor is:

\[ \frac{4!}{(2!)^3 \times 3!} = \frac{24}{8 \times 6} = \frac{24}{48} = \frac{1}{2} \]

Hence, the final answer is:

\[ \text{Final Answer} = 2520 \times \frac{1}{2} = 1260 \]

### Conclusion:
The number of ways to distribute 7 distinct toys among 4 boys such that any 3 boys get 2 toys each and the fourth boy gets 3 toys is **1260**.