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  4. 5 Indian 5 American couples meet at a party shake hands. If no wife shakes hands with her own husband no Indian wife shakes hands with a male, then the number of hand shakes that takes place in the party is -

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

Permutations and Combinations

Solution:

$I _{ H } \rightarrow$ Indian husband, $I _{ w } \rightarrow$ Indian wife
$A _{ H } \rightarrow$ American husband, $A _{ w } \rightarrow$ American wife
Case 1 : Hand shaking occuring between same nationals and same genders (M/F)
$I _{ H }- I _{ H } \rightarrow{ }^5 C _2=10 \text { ways }$
Similarly for $I_w-I_w, A_w-A_w, A_H-A_H$
Total ways $10 4=40$.
Case 2 : All other possible hand shakes
image
Hence total number of handshakes
$=(25 3+40)+(20)=135$
Method-II
Total number of handshakes possible ${ }^{20} C _2$ undesirable handshakes : ${ }^{10} C _2+{ }^{10} C _1$
Hence, Desired ways $={ }^{20} C _2-\left({ }^{10} C _2+{ }^{10} C _1\right)$