TIFR CSE 2019 | Part A | Question: 11

Question

Suppose there are $n$ guests at a party (and no hosts). As the night progresses, the guests meet each other and shake hands. The same pair of guests might shake hands multiple times. for some parties stretch late into the night , and it is hard to keep track.Still, they don’t shake hands with themselves. Let Odd be the set of guests who have shaken an odd number of hands, and let even be the set of guests who have shaken an even number of hands. Which of the following stays invariant throughout the night?

• A. $ \mid \text{Odd} \mid \text{ mod } 2$
• B. $\mid \text{Even} \mid $
• C. $\mid \text{Even} \mid – \mid \text{Odd} \mid$
• D. $2 \mid \text{Even} \mid – \mid \text{Odd} \mid$
• E. $2 \mid \text{Odd} \mid – \mid \text{Even} \mid$

Comments

@srestha What you are telling is correct. But thats not what is asked in the question. 

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Sir

mod2 will not be ans

mod2 just check a number is odd or not

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@srestha

please check my answer.

Answer 1

Answer: Option A

Let O denote the set with guests who have shook odd number of hands and E denote the set with guests who have shook even number of hands at any point of time in the party. Suppose two guests G1 and G2 have just shook hands. We have the following 2 cases

Case1: G1 and G2 are from same set. Suppose they both were in set O before handshake. After handshake, they are both now in set E. |O| = |O|(old) – 2 and |E| = |E|(old) + 2. Same logic for set E.

Case 2: G1 and G2 are from different sets. After shaking hands, they exchange their sets. Suppose G1 is in O and G2 is in E. After handshake G1 is in E and G2 is in O and vice versa. Size of both sets remain unchanged.

In any case, size of a set changes by 2 or remains unchanged. Thus Option A is the answer.