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Bunuel
5 Indian & 5 American couples meet at a party and shake hands. If no wife shakes hands with her own husband and no Indian wife shakes hands with a male then the number of handshakes that takes place in the party is

(A) 95
(B) 110
(C) 135
(D) 150
(E) 160

good cases = all possible cases - bad cases

All possible cases:
From 20 people, the number of ways to form a pair to shake hands = 20C2 \(= \frac{20*19}{2*1} = 190\)

Bad case 1: An Indian woman shakes hands with a male
Number of ways to choose an Indian woman = 5
Number of ways to choose a male = 10
To combine these options, we multiply:
5*10 = 50

Bad case 2: An American woman shakes hands with her own husband
Since there are 5 American couples that could yield a wife shaking hands with her own husband, we get:
5

Thus:
good cases = (all possible cases) - (bad case 1) - (bad case 2) = 190 - 50 - 5 = 135

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

Total nos. of Handshakes = 20C2 = 190

No. of ways when No wives to handshakes with her husbands = 10

No. of ways when IW (Indian wives) doesn't handshakes with any males = 10C2 = 45

So, Required handshakes = 190-10-45 = 135

I think C. :)
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