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5 Indian and 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

Answer Verified Verified
Hint: Find total possible handshakes. Find the number of ways in which a wife shakes hands with husband and Indian wife shakes hands with male. Then subtract these volumes from the total possible handshake.

Complete Step-by-Step solution:
It is said that there are 5 Indian couples and 5 American couples.
Thus total number of person \[=\left( 5\times 2 \right)+\left( 5\times 2 \right)\]
                                                     \[\begin{align}
  & =10+10 \\
 & =20 \\
\end{align}\]
Thus the total possible handshake \[={}^{20}{{C}_{2}}\]
It is said that no wife shakes hands with her own husband and no Indian wife shakes hands with a male.
So let us find the case where a wife shakes hands with husbands and Indian wife shakes hands with male and subtract them from total possible handshakes.
Thus the number of ways in which Indian women shake hands with male, total women = 10, total male = 10.
\[\therefore \] Number of ways Indian women shakes hand with male \[={}^{5}{{C}_{1}}\times {}^{10}{{C}_{1}}\]
                                                                                                         \[\begin{align}
  & =5\times 10 \\
 & =50 \\
\end{align}\]
\[{}^{5}{{C}_{1}}\] is in the form, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[{}^{5}{{C}_{1}}=\dfrac{5!}{1!\left( 5-1 \right)!}=\dfrac{5!}{4!1!}=\dfrac{5\times 4!}{4!1!}=4\]
\[\therefore \] Number of ways Indian women shake hands with makes = 50 ways.
This includes the husbands also.
American women shake hands with their husbands in 5 ways.
\[\therefore \] The desired number of handshakes \[={}^{20}{{C}_{2}}-50-5\]
                                                                       \[\begin{align}
  & =\dfrac{20!}{\left( 20-2 \right)!2!}-50-5 \\
 & =\dfrac{20\times 19\times 18!}{18!2!}-55 \\
 & =\dfrac{20\times 19}{2}-55 \\
 & =190-55 \\
 & =135 \\
\end{align}\]
Thus the number of handshakes in the party = 135.
\[\therefore \] Option (c) is correct.

Note: We use \[{}^{n}{{C}_{r}}\] to find the number of ways to choose r objects from n numbers. Mathematically it is written as \[\dfrac{n!}{r!\left( n-r \right)!}\]. For instance let these be 5 students and we have to choose 2 of them. So we do it in \[{}^{5}{{C}_{2}}\] ways.