89renegade wrote:

Bunuel wrote:

Official Solution:

If all the 10 directors present at a meeting shook hands with each other so that in the end there were no two directors who didn't shake hands, how many handshakes were performed?

A. 30

B. 36

C. 42

D. 45

E. 90

The total number of handshakes will be equal to the number of different pairs possible from these 10 people (one handshake per pair), so \(C^2_{10}=45\).

Answer: D

Hello Bunuel,

I am making a very silly mistake while solving this problem and similar ones which are there in

GMATCLUB tests,

Kindly guide me whats the basic thing I am missing,

1 person does 9 handshakes so 9 people 10 mistakes.

Why is my answer 90 and not 45

I know this s combination problem and should use 10C2 but why is what I am not understanding.

Thanking you in advance.

Hey There,

I would like to give my 2 cents.

Think of these questions in general way.

Lets take example of premier league (for those who are not aware.. it is professional top tier football competition in England)..

There are 20 teams and each teams plays 2 matched with every other team (home and Away).

In total there will be 20 * 19 matches == 380 matches.

Now coming to our question..

Here 10 people are there and so there will be 10 * 9 total handshakes..

But a handshake is same if you look from both persons prospective (i.e, from person 1 or from person 2).

So we have to half it... 10*9 / 2 = 45.

Using this concept we got the formula of 10C2.

Hope it helps.

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