Question

n different toys are to be distributed among n children. Find the number of ways in which these toys can be distributed so that exactly one child gets no toy.

Answer 1

$\frac{n(n-1)n!}{2}$

Answer 2

The problem asks us to find the number of ways to distribute n different toys among n children such that exactly one child gets no toy.

We can solve this problem in two main steps:

Step 1: Choose the child who gets no toy.

There are n children in total. We need to choose exactly one child who will not receive any toy. The number of ways to choose this child is given by the combination formula:

$n C 1 = \frac{n!}{(1! * (n-1)!)} = n$ ways.

Step 2: Distribute the n different toys among the remaining n-1 children such that each of these n-1 children receives at least one toy.

Since there are n toys and n-1 children, and each child must receive at least one toy, it logically follows that one child among these n-1 children must receive two toys, and the remaining n-2 children must each receive one toy.

To achieve this distribution, we can follow these sub-steps:

a. Choose 2 toys out of n different toys to be given to one specific child. The number of ways to choose these 2 toys is $n C 2 = \frac{n!}{(2! * (n-2)!}$.

b. Choose 1 child out of the n-1 children (who are receiving toys) to receive these 2 toys. The number of ways to choose this child is $(n-1) C 1 = (n-1)$.

c. Distribute the remaining n-2 toys among the remaining n-2 children. Each of these n-2 children will receive exactly one toy. Since both the toys and children are distinct, this is a permutation problem. The number of ways to distribute these n-2 distinct toys to n-2 distinct children is $(n-2)!$.

Now, multiply the results of these sub-steps to find the number of ways for Step 2: Number of ways for Step 2 = $n C 2 * (n-1) C 1 * (n-2)!$ $= [\frac{n!}{(2! * (n-2)!)}] * (n-1) * (n-2)!$

We can cancel $(n-2)!$ from the numerator and denominator:

$= [\frac{n!}{2!}] * (n-1)$ $= \frac{n! * (n-1)}{2}$

Step 3: Combine the results from Step 1 and Step 2.

The total number of ways to distribute the toys is the product of the ways from Step 1 and Step 2: Total ways = (Ways to choose the child who gets no toy) * (Ways to distribute toys to the remaining children) Total ways = $n * [\frac{n! * (n-1)}{2}]$ Total ways = $\frac{n * (n-1) * n!}{2}$

The final answer is $\frac{n(n-1)n!}{2}$.

Explanation of the solution:

1. Select the child who receives no toy: There are n choices for this child (n C 1).
2. Distribute n distinct toys among the remaining n-1 distinct children such that each gets at least one toy:
   • Since there are n toys and n-1 children, one child must receive two toys, and the other n-2 children must receive one toy each.
   • Choose 2 toys out of n to form a pair (n C 2 ways).
   • Choose 1 child out of the n-1 children to receive this pair (n-1 C 1 ways).
   • Distribute the remaining n-2 toys among the remaining n-2 children, one toy each ((n-2)! ways).
   • Multiply these counts: $n C 2 * (n-1) C 1 * (n-2)! = [\frac{n!}{(2!(n-2)!)}] * (n-1) * (n-2)! = \frac{n! * (n-1)}{2}$.
3. Total ways: Multiply the result from step 1 and step 2: $n * [\frac{n! * (n-1)}{2}] = \frac{n(n-1)n!}{2}$.