500 ps test6 #20

Hi, This is the sum of a progression kind of problem.

The 1st person would shake with 9 others, the 2nd would do with 8 others etc. So, the final answer would 9+8+...+1 = 9*(9+1)/2 = 45.
Answer D.

Simply put, we are asked to find the number of ways 10 people can
shake hands with each other.

10C2 is the answer

Thank you.

I am always confursed using P and C... (if you could a bit explain, that would be soo great...)

In this case, 10C2, 10 is obviously 10 people, but where the number 2 come from? meaning avoiding duplication?

Thanks!

dunkin77 wrote:Thank you.

I am always confursed using P and C... (if you could a bit explain, that would be soo great...)

In this case, 10C2, 10 is obviously 10 people, but where the number 2 come from? meaning avoiding duplication?

Thanks!

You might want to check out the GMAT resource wiki. It has a lot of
useful info in there that I have personally found helpful

https://www.beatthegmat.com/wiki/doku.ph ... _resources

If 10 persons meet at a reunion and each person shakes hands exactly once with each of the others, what is the total number of handshakes?
(A) 10•9•8•7•6•5•4•3•2•1
(B) 10•10
(C) 10•9
(D) 45
(E) 36

For a handshake 2 people are required. Hence 2 people can be selected from a group of 10 in 10C2 ways

= 45

Hence Choice D

since each person shakes hands exactly once with each of the others, this is just the same as grouping two people together without considering the order.
e.g in {a,b,c,d}, the number of handshakes will be 6. i.e {a,b], {a,c}, {a,d}, {b,c}, {b,d} and {c,d}.
This is given by $$n_{Cr}$$ , where n is the total number of people and r is the number of people shaking hands once.
Total number of handshakes is thus given as
$$_{10_{C2}=\ \frac{10!}{\left(10-2\right)!2!}\ =\ \frac{\left(10\cdot9\cdot8!\right)}{8!\cdot2!}=\frac{\left(10\cdot9\right)}{2}=45}$$
$$Therefore,\ there\ will\ be\ 45\ handshakes\ if\ two\ people\ do\ not\ shake\ hands\ with\ each\ other\ more\ than\ one\ time$$