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Bunuel
At the beginning of a party, each person present shook hands with all other people present and there were in all 28 handshakes. In the midst of the party,2 persons left due to an emergency. Now, the number of males and females present in the party was equal. At the end, each female shook hands only with every female present and each male shook hands only with every male present. What is the total number of handshakes that took place at the party?

(A) 52
(B) 45
(C) 35
(D) 34
(E) 33

Solution:

Let n be the number of people originally at the party. Since there were 28 handshakes exchanged, we can create the equation:

nC2 = 28

n(n - 1) / 2 = 28

n(n - 1) = 56

Without solving the equation any further, we can see that n must be 8 since 8 x 7 = 56. Now, since 2 people have left and there are an equal number of men and women, there must be 3 men and 3 women. Since the men shake hands only with men, there are 3C2 = 3 handshakes between the men. Likewise, since the women shake hands only with women, there are also 3C2 = 3 handshakes between the women. Therefore, including the first 28 handshakes, there are a total of 28 + 3 + 3 = 34 handshakes.

Answer: D


I agree to all the numerical values calc..
But in the end the task of Women shaking hands with women AND men shaking hand with men.

As per my learnings..

since the two tasks are necessary tasks and not optional tasks, then should we not use X (Multiplication sign) between 3 & 3 instead of + sign?
Or am I confusing my self with this task related concept?
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