Question

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

Answer 1

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

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=\[^{20}{C_2}\]
\[
   \Rightarrow \dfrac{{20!}}{{(20 - 2)!2!}} \\
   \Rightarrow \dfrac{{20!}}{{18!2!}} = \dfrac{{20 \times 19 \times 18!}}{{18! \times 2}} = 190 \\
\]
Indian women handshakes with male=\[^5{C_1}{ \times ^{10}}{C_1}\]
\[
   \Rightarrow \dfrac{{5!}}{{(5 - 1)!1!}} \times \dfrac{{10!}}{{(10 - 1)!1!}} \\
   \Rightarrow \dfrac{{5!}}{{4!}} \times \dfrac{{10!}}{{9!}} = \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{10 \times 9!}}{{9!}} = 5 \times 10 = 50 \\
\]
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.
\[ \Rightarrow \]Option (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.