Question

The number of ways in which a mixed double game can be arranged from amongst 9 married couples if no husband and wife play in the same game is:
(A). 756
(B). 1512
(C). 3024
(D). None of these

Answer 1

Hint: Assume different variables for man and woman. Now for the game we need 2 men, 2 women. By using the definition of combination, try to find a number of ways to select 2 men. Now exclude these 2 couples. From remaining couples select 2 women players. Now multiply these 2 numbers. In each selection 1 man has 2 options of women to pair up. So, multiply by 2. Those will be the final result.

Complete step-by-step answer:
It is a mathematical technique that determines the number of possible arrangements in a collection of items where the order of selection does not matter in combination you can select items in any order. Formula is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Mixed double tennis game:- It is a tennis game between 2 teams in which each team has 1 man, 1 woman as players.
Given that there are 9 couples who are to be selected.
Let us assume 9 men as a, b, c, d,……….. i
Let us assume 9 women as A, B, C, D,………….. I
Factorial:- In mathematics, factorial is an operation, denoted by “$\left( ! \right)$ “. It represents the product of all numbers between 1 and a given number.
In simple words factorial of a number can be found by multiplying all terms you get by subtracting 1 from a given number repeatedly till you get the number difference as 1. Its representation can be written as: $n!=n\times \left( n-1 \right)\times \left( n-2 \right)...................\times 1$
For example: $5!=5\times 4\times 3\times 2\times 1=120.$
Note that we assume $0!=1$ . It is standard value. It has a wide range of applications in combinations.
Now, for the game we need 2 men. So number of ways: $Men={}^{9}{{C}_{2}}$ .
Now we need 2 women, to complete solutions for the game.
Given the question that no couple plays a game at a time. So, we must remove 2 women who are wives of 2 selected men.
Now we are left with 7 choices. So, $women={}^{7}{{C}_{2}}$ .
Let us assume in Men a, e are selected and in women B, D are selected the possible matches can be between: $\left( a,B \right)v/s\left( e,D \right)$ and $\left( a,D \right)v/s\left( e,B \right)$ .
So, each man has 2 possibilities to select. So, 2 matches can be found. So, we need to multiply by 2.
So, Total number of ways $={}^{9}{{C}_{2}}{}^{7}{{C}_{2}}2$
By substituting the values, we get value as $=\dfrac{9!}{7!2!}\dfrac{7!}{5!2!}2!$
By simplifying the values, we get it as $=3\times 7\times 8\times 9$ .
Finally the number of ways is 1512 ways.

Therefore, Option (B) is correct for the given equation.

Note: Generally students get confused between combinations, permutation if you have to select a combination, If you have to arrange then use permutation. It is very nice trick to use
Don’t forget to consider all possibilities or else you might reach the wrong answer. Do it carefully for example: here if you multiply by 2 then you might get the wrong answer.