Question

Mixed double tennis are conducted among $9$ couples. Name of possible matches such that there is no couple in any $SIDE$ are__

• A. $9_{C_2}{7}_{P_2}$
• B. $9_{C_2}\left(23\right)$
• C. $9_{C_2}\left(57\right)$
• D. $9_{C_2}\left(42\right)$

Answer 1

The correct option is C $9_{C_2}\left(57\right)$

$2$ men are selected from $9$ men in ${}^9{C}_2$ ways.

Since no couple in any side.

Therefore,

i)Two women out of the remaining $7$ are chosen in ${}^7{C}_2$ ways.

Now one team can be chosen as $(M_1,W_1)$ or $(M_1,W_2)$ in $2$ ways.

ii) One from the 2 women and One from the remaning 7

Here only one combination is possible so that they don't form a couple

iii)Two from the 2 women

Here only one combination is possible so that they don't form a couple

Therefore,

Required no. of arrangements $={}^9{C}_2({}^7{C}_2\times 2+{}^2{C}_1\times {}^7{C}_1\times 1+{}^2{C}_2\times 1)={}^9{C}_2(21\times 2+14+1)={}^9{C}_2\times 57$