Question

In how many ways a mixed doubles game can be arranged from 8 married couples if no husband and wife play in the same game?

Answer 1

Consider the case where first we select 2 husbands out of the 8 available and that will be given by ${}^{\text{8}}{{\text{C}}_2}$ ways , Now since no husband and wife play in the same game total wifes available for selection from will be 6 . We can do this by selective wifes first as well so multiply with 2 .

Complete step by step answer :
As given in the question,
We have to find the number of ways in which 8 couples can be organized, So, we have to use "combinations"
2 gents can be selected from 8 gents in ${}^8{{\text{C}}_2}$ ways
2 ladies can be selected from 8−2=6 ladies
in ${}^6{{\text{C}}_2}$ ways
∴ The number of required ways = ${}^8{{\text{C}}_2}$ × ${}^6{{\text{C}}_2}$ ×2
= $\dfrac{{8!}}{{{\text{[2}}!{\text{(8 - 2)}}!{\text{]}}}}$× $\dfrac{{6!}}{{{\text{[2}}!{\text{(6 - 2)}}!{\text{]}}}}$ ×2
= $\dfrac{{8!}}{{{\text{[2}}!{\text{(6)}}!{\text{]}}}}$× $\dfrac{{6!}}{{{\text{[2}}!{\text{(4)}}!{\text{]}}}}$×2
= $\dfrac{{(6!) \times 7 \times 8}}{{{\text{[2}}!{\text{(6)}}!{\text{]}}}}$× $\dfrac{{(4!) \times 5 \times 6}}{{{\text{[2}}!{\text{(4)}}!{\text{]}}}}$×2
= $\dfrac{{7 \times 8}}{{{\text{[2}}!{\text{]}}}}$× $\dfrac{{5 \times 6}}{{{\text{[2}}!{\text{]}}}}$×2
=28 × 15 ×2
= 840
Thus, in total 840 ways a mixed doubles game can be arranged from 8 married no husband and wife play in the same game.

Note: Combinations are the way of selecting the objects or numbers from a group of objects or collection, in such a way that the order of the objects does not matter. For example, suppose we have a set of three letters: A, B, and C. Each possible selection would be an example of a combination.
And also, we need to remember this formula for selecting r things out of n.
${}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n!}}}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}$
where n! means the factorial of n.
for example, $3!{\text{ = 3}} \times {\text{2}} \times 1$