First slide

Permutations

Question

The number of ways in which a mixed doubles game can be arranged from amongst n couples such that no husband and wife play in the same game, is

Moderate

A

$^{n} P_{4}$
B

$^{n} C_{4}$
C

$\frac{1}{2} {}^{n}P_{4}$
D

$\frac{1}{2} {}^{n}C_{4}$

Solution

A mixed doubles game involves two males and two females. Two males can be chosen from n males in ⁿC₂ ways. Having chosen two males, now 2 females are to be chosen from (n –2) females leaving the wives of the two already chosen males. This can be done in ^{n –2}C₂ ways.
Hence, the required number of ways
$=^{n} C_{2} \cdot^{n - 2} C_{2} \cdot 2$ [for every choice of 2 males and 2 females, they can be paired in 2 ways].
$\begin{array}{l} = \frac{n (n - 1)}{2} \cdot \frac{(n - 2) (n - 3)}{2} \cdot 2 \\ = \frac{n (n - 1) (n - 2) (n - 3)}{2} = \frac{1}{2} \cdot^{n} P_{4} \end{array}$

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