Question

Number of ways in which 5 different toys can be distributed among 5 children if exactly one child do not get any toy is:
\[\begin{align}
  & A.\text{ }1100 \\
 & B.\text{ }1200 \\
 & C.\text{ }1300 \\
 & D.\text{ }240 \\
\end{align}\]

Answer 1

Hint: In this question, we will use a combination method to get our required answer. First we will have to understand the logic behind this and then solve further. We will consider all possibilities, such as, ways of one child not getting any toy, 5 toys left for four children which means one child will get two toys and distributing remaining 3 toys to 3 children. After that, we will multiply all the possibilities to conclude our final answer. Formula for combination that we will use is \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\] where we have to choose r things from a set of n things without replacement and here order does not matter.

Complete step-by-step answer:
Let us understand the combination first. Combination is a way of selecting items from a collection, such that the order of selection does not matter. Let us now analyze the solution.
We are given five toys and five children. But one of them does not get any toy. Hence, total ways in which a child does not get toys becomes 5.
Number of ways child does not get toy = 5 . . . . . . . . . . . . . . . . (1)
Now, four children are left with 5 toys. Therefore, one of them will get 2 toys.
Hence, any two toys from five toys can be selected for that one child. Therefore, ways become ${}^{5}{{C}_{2}}$ where 5 is the number of toys out of which 2 toys will be chosen.
\[{}^{5}{{C}_{2}}=\dfrac{5!}{2!\left( 5-2 \right)!}=\dfrac{5!}{2!3!}=\dfrac{5\times 4\times 3\times 2\times 1}{2\times 3\times 2}=10\]
Number of ways two toys are selected = 10 . . . . . . . . . . . . . . . . . (2)
Now, four of them can get those two toys. Hence,
Number of ways second child gets two toys = 4 . . . . . . . . . . . . . . . (3)
Now, we are left with three toys and three children.
So, if we get one of the three toys to one child,
Ways in which third child get any of toy = 3 . . . . . . . . . . . . . . . . . . . (4)
Now we are left with two toys and two children. Therefore,
Ways in which fourth child gets toy = 2 . . . . . . . . . . . . . . . . . (5)
And at last, one toy is left for one child and hence,
Ways in which fifth child gets toy = 1 . . . . . . . . . . . . . . . . (6)
Combining all ways we get:
Total ways \[\Rightarrow 5\times 10\times 4\times 3\times 2\times 1=1200\]
Hence, option B is the correct answer.

So, the correct answer is “Option B”.

Note: These sums are very complicated and students should not forget any possibility. Ways of selecting toys and ways of giving them to children both should be considered. At last, students should always multiply all the ways as they will happen simultaneously. Multiplication is used when we use 'and' and addition is used when we use 'or' for our selections.