Question

Nine toys are to be packed in nine boxes. If 5 of them are too big for 3 boxes, then the number of ways in which they can be packed is –
A) $^6{P_5} \times 4!$
B) $^6{P_5} \times 3!$
C) $6! \times 3! \times 2!$
D) $5! \times 3!$

Answer 1

Hint: In this question, we are given nine toys to be packed in 9 boxes. But out of these 9 toys and 9 boxes, some of them are big whereas others are small. First, arrange the big toys in big boxes. Then pack the smaller toys in remaining big and small boxes.

Complete step-by-step solution:
We are given nine toys and we have been asked to pack them in nine boxes. But the question says that – 5 toys out of the 9 toys are too big for 3 boxes. The inference of this line can be stated below –
Out of 9 toys, 5 toys are big whereas 4 toys are small.
Also, 3 boxes are small and 6 are big.
Our 5 big toys can settle in 5 of the 6 big boxes only but 4 small toys can settle in both big and small boxes. So, first we will arrange 5 big boxes and then we will arrange the remaining 4 toys in the remaining number of boxes.
Step 1: Arrange 5 big toys in 6 boxes first.
This can be done in $^6{P_5}$ ways.
Step 2: Now, we have 4 boxes left. Out of these boxes, 1 box is big whereas 3 other boxes are small. And we have 4 small toys left. Small toys can fit in big boxes as well. Now, arrange 4 toys in these 4 boxes.
This can be done in $4!$ ways.

$\therefore $ This arrangement can be made in option (A) $^6{P_5} \times 4!$ ways.

Note: In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The work permutation also refers to the act or process of changing the linear order of an ordered set. Permutation differs from combinations, which are selections of some members of a set regardless of order.