Question

The number of ways in which $3$ friends can stay in two hotels is:
(A) ${2^3}$
(B) ${3^2}$
(C) $^3{P_2}$
(D) None of these

Answer 1

Hint: In the given question, we have to find the number of ways in which three friends can be allocated to two hotels. So, we will first find the number of options of hotels with each friend. Then, we will use the multiplication rule of counting by multiplying the number of options with each person to find the total number of ways in which the three friends can stay in the hotels.

Complete step-by-step solution:
So, we are to find the number of ways in which the three friends can stay in the two hotels.
So, total number of hotels $ = 2$
Number of friends $ = 3$
Number of options of hotels that can be chosen by first friend $ = 2$
Similarly, the remaining two friends also have $2$ options of hotels to choose from.
Now, according to the multiplication rule of the counting, if there are m ways of doing one thing and n ways of doing another thing, then there are a total $m \times n$ ways of doing both the things simultaneously.
So, total number of ways in which $3$ friends can stay in two hotels $ = 2 \times 2 \times 2$
$ = {2^3}$
Computing the power, we get,
$ = 8$
So, there are a total of $8$ ways in which the three friends can stay in two hotels.
So, option (A) is the correct answer.

Note: In such a problem, one must know the multiplication rule of counting as consequent events are occurring. One must take care while doing the calculations and should recheck them so as to be sure of the final answer. We should give emphasis on language as it changes the entire problem and method of solution as well.