The number of ways we can choose two teams of mixed double for a tennis tournament from four couples such that in any couple participates, then it is in same team is

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None of these

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Solution

The correct option is B 42
Two team double can formed as:
$\begin{matrix} C a s e 1 : W h e n n o c o u p l e i s c h o s e n = 2 \times^{4} C_{2} w a y s C a s e 2 : W h e n o n l y o n e c o u p l e i s c h o s e n =^{4} C_{1} \times^{3} C_{1} \times^{2} C_{1} C a s e 3 : W h e n t w o c o u p l e i s c h o s e n =^{4} C_{2} w a y s H e n c e, t o t a l n u m b e r o f w a y s i s : = 2 \times^{4} C_{2} +^{4} C_{1} \times^{3} C_{1} \times^{2} C +^{4} C_{2} = 42 \end{matrix}$

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