In how many ways can 20 identical toys be distributed among 4 children so that each one gets at least 3 toys?

Options

A. `""^(11)C_(2)`
B. `""^(12)C_(3)`
C. `""^(11)C_(3)`
D. `""^(12)C_(4)`

Explanation

To solve the problem of distributing 20 identical toys among 4 children such that each child receives at least 3 toys, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Requirement**: Each of the 4 children must receive at least 3 toys. Therefore, we first allocate 3 toys to each child.

2. **Allocate Initial Toys**:
- Since there are 4 children, the total number of toys given initially is:
\[ 3 \text{ toys/child} \times 4 \text{ children} = 12 \text{ toys} \]
- After giving out 12 toys, we have:
\[ 20 \text{ total toys} - 12 \text{ toys given} = 8 \text{ toys remaining} \]

3. **Reformulate the Problem**: Now, we need to distribute the remaining 8 toys among the 4 children with no restrictions (since each child has already received their minimum of 3 toys).

4. **Use the Stars and Bars Theorem**: This is a combinatorial method used to find the number of ways to distribute \( n \) identical items into \( r \) distinct groups. The formula for this is:
\[ \text{Number of ways} = \binom{n + r - 1}{r - 1} \]
where \( n \) is the number of identical items (toys), and \( r \) is the number of groups (children).

5. **Substitute Values**: In our case, we have:
- \( n = 8 \) (remaining toys)
- \( r = 4 \) (children)
- Therefore, we substitute into the formula:
\[ \text{Number of ways} = \binom{8 + 4 - 1}{4 - 1} = \binom{11}{3} \]

6. **Calculate the Combination**:
- We calculate \( \binom{11}{3} \):
\[ \binom{11}{3} = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = \frac{990}{6} = 165 \]

7. **Final Answer**: Thus, the total number of ways to distribute the 20 identical toys among 4 children, ensuring each child gets at least 3 toys, is:
\[ \boxed{165} \]