Question

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

Answer 1

135

Answer 2

The problem involves calculating the number of handshakes under specific restrictions.

1. Total number of people: There are 5 Indian couples and 5 American couples, so there are 10 couples in total. This means there are 10 males and 10 females, making a total of 20 people.

2. Total possible handshakes without any restrictions: The number of handshakes among N people is given by the combination formula $^N C_2$. For 20 people, the total possible handshakes are: $^{20}C_2 = \frac{20 \times (20-1)}{2 \times 1} = \frac{20 \times 19}{2} = 10 \times 19 = 190$

3. Identify and quantify forbidden handshakes based on the restrictions:

• Restriction 1: "no wife shakes hands with her husband" There are 10 couples. In each couple, the wife and husband are forbidden from shaking hands. Number of forbidden handshakes due to this restriction = 10. Let this set of forbidden handshakes be $F_1$. So, $|F_1| = 10$.

• Restriction 2: "no Indian wife shakes hands with a male" There are 5 Indian wives. There are 10 males in total (5 Indian males + 5 American males). Each Indian wife is forbidden from shaking hands with any male. Number of forbidden handshakes due to this restriction = 5 (Indian wives) $\times$ 10 (males) = 50. Let this set of forbidden handshakes be $F_2$. So, $|F_2| = 50$.

4. Account for overlaps (double-counted forbidden handshakes): We need to find the number of handshakes that are forbidden by both restrictions, i.e., $F_1 \cap F_2$. A handshake is in $F_1 \cap F_2$ if it is between a wife and her husband AND it is between an Indian wife and a male. This means the handshake must be between an Indian wife and her Indian husband. There are 5 Indian couples, so there are 5 such handshakes (Indian wife with her Indian husband). So, $|F_1 \cap F_2| = 5$.

5. Calculate the total number of unique forbidden handshakes: Using the principle of inclusion-exclusion, the total number of unique forbidden handshakes ($F_1 \cup F_2$) is: $|F_1 \cup F_2| = |F_1| + |F_2| - |F_1 \cap F_2|$ $|F_1 \cup F_2| = 10 + 50 - 5 = 55$

6. Calculate the number of allowed handshakes: Number of allowed handshakes = Total possible handshakes - Total unique forbidden handshakes $= 190 - 55 = 135$

The number of handshakes that take place in the party is 135.

Explanation of the solution:

1. Calculate total possible handshakes: $C(20, 2) = 190$.
2. Identify forbidden handshakes: a. Husband-wife pairs: 10 couples $\implies$ 10 forbidden handshakes. b. Indian wife-male pairs: 5 Indian wives * 10 males $\implies$ 50 forbidden handshakes.
3. Identify overlaps: Handshakes between Indian wives and their own husbands are counted in both categories (5 such handshakes).
4. Total unique forbidden handshakes = (Forbidden from a) + (Forbidden from b) - (Overlaps) = $10 + 50 - 5 = 55$.
5. Allowed handshakes = Total possible - Total unique forbidden = $190 - 55 = 135$.

Answer: 135