Question

The number of ways in which six different prizes can be distributed among three children each receiving at least one prize is :
A. 270
B. 540
C. 1080
D. 2160

Answer 1

Hint: The prizes can be distributed in only three types of patterns. In the first type, it can be 1 prize, 2 prizes, and 3 prizes to three children. In the second pattern, it can be 2 prizes to each child and the third pattern can have 4 prizes to one child and 1 to each of the remaining children. So, we will find the number of ways for each of these patterns and hence find the number of ways of distributing the prizes.

Complete step by step answer:
In the question, we are asked to find the number of ways of distributing six different prizes among three children such that each child receives at least one prize. Now, we know that the only way to distribute the 6 prizes among 6 children can be done in only 3 ways. We can do it by giving 1 prize, 2 prizes and 3 prizes to the respective three children or by giving 2 prizes to each child or by giving 4 prizes to one child and 1 to each to two children. So, we will now consider each of the way and find the number of ways of distributing to satisfy the three ways and then find their sum to get the total number of ways.
So, the first way of distribution, we will be giving, 1 prize, 2 prizes and 3 prizes to the respective children, So, the number of ways will be given by
$\begin{align}
  & \dfrac{6!}{3!\times 2!\times 1!}\times 3! \\
 & \Rightarrow \dfrac{720}{6\times 2\times 1}\times 6 \\
 & \Rightarrow 360 \\
\end{align}$
So, we get the number of ways as 360.
Now let us consider the second case, where we give 2 prizes to each child. So, we get,
$\begin{align}
  & \dfrac{6!}{2!\times 2!\times 2!\times 3!}\times 3! \\
 & \Rightarrow \dfrac{720}{2\times 2\times 2\times 6}\times 6 \\
 & \Rightarrow 90 \\
\end{align}$
So, we get 90 ways this way.
Now, let us consider the last way, that is giving 4 prizes to one child and 1 to each to two children. So, we get,
$\begin{align}
  & \dfrac{6!}{4!\times 2!}\times 3! \\
 & \Rightarrow \dfrac{720}{24\times 2}\times 6 \\
 & \Rightarrow 90 \\
\end{align}$
So, we get 90 ways again.
Thus, we get the total number of ways as (360 + 90 + 90) = 540 ways.
Therefore, the correct answer is option B.

Note:
In combinatorics, the rule of sum or addition principle states that it is the idea that if the complete A ways of doing something and B ways of doing something and B ways of doing something, we cannot do at the same time, then there are (A + B) ways of choosing one of the actions. In this question, there is a possibility that the student might miss considering all the cases. They may forget to consider case 2 and get a different answer.