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There are $5$ people in the room

No. of people involved in one handshake $= 2$

So, we will use combination over here to find the no. of possible handshakes $^nC_r = \frac{n!}{(n-r)!r!}.$

Here$,n = 5,r = 2$

Now, $^5C_2 = \frac{5!}{3!2!} = \frac{5.4.3!}{3!.2.1} = 10$

Hence the maximum number of distinct handshakes that can happen in the room with $5$ people in it is $10.$

So, the correct answer is $(B).$

No. of people involved in one handshake $= 2$

So, we will use combination over here to find the no. of possible handshakes $^nC_r = \frac{n!}{(n-r)!r!}.$

Here$,n = 5,r = 2$

Now, $^5C_2 = \frac{5!}{3!2!} = \frac{5.4.3!}{3!.2.1} = 10$

Hence the maximum number of distinct handshakes that can happen in the room with $5$ people in it is $10.$

So, the correct answer is $(B).$