Number of ways in which 5 different toys can be distributed in 5 children if exactly one child does not get any toy is greater than

Options

A. 1200
B. 2400
C. 960
D. 300

Explanation

To solve the problem of distributing 5 different toys among 5 children such that exactly one child does not receive any toy, we can break down the solution into the following steps:

### Step 1: Choose the child who will not receive a toy
We have 5 children, and we need to choose 1 child who will not receive any toy. The number of ways to choose 1 child from 5 is given by:

\[ \text{Ways to choose 1 child} = 5 \]

### Step 2: Distribute the toys among the remaining children
After choosing 1 child who will not receive a toy, we are left with 4 children. We need to distribute 5 different toys among these 4 children, with the condition that at least one of the children must receive 2 toys.

### Step 3: Choose the child who will receive 2 toys
We can choose 1 child from the 4 remaining children to receive 2 toys. The number of ways to choose this child is:

\[ \text{Ways to choose 1 child from 4} = 4 \]

### Step 4: Choose 2 toys for that child
Now, we need to select 2 toys out of the 5 to give to the chosen child. The number of ways to choose 2 toys from 5 is given by the combination formula:

\[ \text{Ways to choose 2 toys from 5} = \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Step 5: Distribute the remaining toys
After giving 2 toys to one child, we have 3 toys left to distribute among the remaining 3 children. Each of these children will receive exactly 1 toy. The number of ways to distribute 3 different toys to 3 different children is given by the factorial of the number of toys (or children):

\[ \text{Ways to distribute 3 toys to 3 children} = 3! = 6 \]

### Step 6: Calculate the total number of ways
Now, we can combine all the steps to find the total number of ways to distribute the toys:

\[ \text{Total ways} = (\text{Ways to choose 1 child}) \times (\text{Ways to choose 1 child from 4}) \times (\text{Ways to choose 2 toys}) \times (\text{Ways to distribute remaining toys}) \]

Substituting the values we calculated:

\[ \text{Total ways} = 5 \times 4 \times 10 \times 6 \]

Calculating this gives:

\[ \text{Total ways} = 5 \times 4 = 20 \]
\[ 20 \times 10 = 200 \]
\[ 200 \times 6 = 1200 \]

Thus, the total number of ways to distribute the 5 different toys among 5 children, with exactly one child not receiving any toy, is **1200**.

### Final Answer:
**1200**

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