Question

Two friends A and B have an equal number of daughters. There are three cinema tickets which are to be distributed among the daughters of A and B. The probability that all the tickets go to daughters of A is $\dfrac{1}{20}.$ The number of daughters each of them have is
A.4
B.3
C.6
D.1

Answer 1

We are required to find the number of daughters each of the two friends A and B have. This can be done by using the concept of permutations and combinations. In this case we use combinations since order does not matter here. We consider the total number of daughters to be 2n, where n stands for the number of daughters A or B has. The total number of ways in which these 3 tickets can go to all of them needs to be calculated. Then we need to calculate the number of ways in which only the daughters of A receive them. We divide them and equate it to the probability value $\dfrac{1}{20}$ and solve for n.

Complete step by step answer:
To solve this question, let us consider both A and B to have n number of daughters. This is so because according to the question, both of them have an equal number of daughters. Now let us consider the total number of ways in which the 3 tickets can be distributed among all the daughters which is 2n. Since order does not matter, we use combinations. Therefore,
$\Rightarrow \text{Total no}\text{. of ways=}{}^{2n}{{C}_{3}}$
The above term represents the total number of ways in which 3 tickets can be distributed among all the daughters of A and B which are 2n.
Now, we will consider the case in which the three tickets are distributed among the daughters of A. We need to distribute 3 tickets in n ways.
$\Rightarrow \text{Favourable no}\text{. of ways=}{}^{n}{{C}_{3}}$
Now, the probability of all tickets going to daughters of A is given as $\dfrac{1}{20}.$ This is same as the favourable number of ways divided by the total number of ways.
$\Rightarrow \text{Probability=}\dfrac{\text{Favourable no}\text{. of ways}}{\text{Total no}\text{. of ways}}$
Substituting the values,
$\Rightarrow \text{Probability=}\dfrac{{}^{n}{{C}_{3}}}{{}^{2n}{{C}_{3}}}$
Combinations can be given by the formula ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.$
Substituting the value of probability and formula for combinations,
$\Rightarrow \dfrac{1}{20}\text{=}\dfrac{{}^{n}{{C}_{3}}}{{}^{2n}{{C}_{3}}}$
$\Rightarrow \dfrac{\dfrac{n!}{3!\left( n-3 \right)!}}{\dfrac{2n!}{3!\left( 2n-3 \right)!}}=\dfrac{1}{20}$
Expanding the factorials,
$\Rightarrow \dfrac{\dfrac{n.\left( n-1 \right).\left( n-2 \right).\left( n-3 \right)!}{3!\left( n-3 \right)!}}{\dfrac{2n.\left( 2n-1 \right).\left( 2n-2 \right).\left( 2n-3 \right)!}{3!\left( 2n-3 \right)!}}=\dfrac{1}{20}$
Cancelling the factorial terms that are same and cancelling the $3!$ denominator of the numerator term and the $3!$ denominator of the denominator,
$\Rightarrow \dfrac{n.\left( n-1 \right).\left( n-2 \right)}{2n.\left( 2n-1 \right).\left( 2n-2 \right)}=\dfrac{1}{20}$
Taking 2 common out from the last term of the denominator and cancelling the n in the numerator and denominator,
$\Rightarrow \dfrac{\left( n-1 \right).\left( n-2 \right)}{2.\left( 2n-1 \right).2.\left( n-1 \right)}=\dfrac{1}{20}$
Cancelling the $\left( n-1 \right)$ terms in the numerator and denominator,
$\Rightarrow \dfrac{\left( n-2 \right)}{4\left( 2n-1 \right)}=\dfrac{1}{20}$
Cross multiplying both sides,
$\Rightarrow 20.\left( n-2 \right)=4\left( 2n-1 \right)$
Expanding both sides,
$\Rightarrow 20n-40=8n-4$
Taking all n terms to one side and constants to other side,
$\Rightarrow 20n-8n=40-4$
Simplifying this,
$\Rightarrow 12n=36$
Dividing both sides by 12,
$\Rightarrow n=3$

So, the correct answer is “Option B”.

Note: It is important to know the basics of permutations and combinations to solve such questions. It is necessary to note that we use permutations when order of arrangement matters. In the other case, when the order of arrangement does not matter, we use combinations as shown in this problem. Here, the order in which the tickets are given to the daughters do not matter. That is why we use combinations.