5 Indians & 5 american couples meet at a party & shake hands. If no wife shakes hands with her own husband & Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party is

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110

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135

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150

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Solution

The correct option is C 135
There are $10$ male and $10$ female.
Total hand-shake $=^{20} C_{2}$
$= 190$ .
Indian women hand-shake with male $=^{5} C_{1} \times^{10} C_{1} = 50$
(It includes handshake with her husband also)
American women hand-shake with their own husband $= 5$ ways.
So, required hand-shake $= 190 - 50 - 5$
$= 135$

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5 Indians & 5 american couples meet at a party & shake hands. If no wife shakes hands with her own husband & Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party is

The correct option is C 135There are 10 male and 10 female.Total hand-shake = 20C2 =190.Indian women hand-shake with male = 5C1× 10C1=50(It includes handshake with her husband also)American women hand-shake with their own husband =5 ways.So, required hand-shake =190−50−5 =135