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Question: 5 Indian & 5 American couples meet at a party and shake hands. If no wife shakes hands with her husb...

5 Indian & 5 American couples meet at a party and shake hands. If no wife shakes hands with her husband and no Indian wife shakes hand with a male, then the number of handshakes that takes place in the party is ‘m’ then;
(A) m=135
(B) m is divisible by 5
(C) m is divisible by 7
(D) m is divisible by 3

Explanation

Solution

The number of way to choose a sample of r elements from a set of n distinct objects where order does not matter is=nCr=n!(nr)!r!{ = ^n}{C_r} = \dfrac{{n!}}{{(n - r)!r!}}

Complete step-by-step answer:
Given: 5 Indian couples and 5 American couples.
There are 10 males and 10 females present in the party.
Total handshakes=20C2^{20}{C_2}

20!(202)!2! 20!18!2!=20×19×18!18!×2=190  \Rightarrow \dfrac{{20!}}{{(20 - 2)!2!}} \\\ \Rightarrow \dfrac{{20!}}{{18!2!}} = \dfrac{{20 \times 19 \times 18!}}{{18! \times 2}} = 190 \\\

Indian women handshakes with male=5C1×10C1^5{C_1}{ \times ^{10}}{C_1}

5!(51)!1!×10!(101)!1! 5!4!×10!9!=5×4!4!×10×9!9!=5×10=50  \Rightarrow \dfrac{{5!}}{{(5 - 1)!1!}} \times \dfrac{{10!}}{{(10 - 1)!1!}} \\\ \Rightarrow \dfrac{{5!}}{{4!}} \times \dfrac{{10!}}{{9!}} = \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{10 \times 9!}}{{9!}} = 5 \times 10 = 50 \\\

It includes the handshakes with her husband too.
American women handshakes with their own husband= 5 ways
So, required handshakes=190-50-5=135 ways
Since, 135 is divided by 5 and 3.
\RightarrowOption (A), (B) and (D) is correct.

Note: Students should know the difference between permutation and combination. Permutations refers to arrangements whereas combinations refers to selection. Also, go through the conditions carefully in the question.