Question

If you have six people in a room, and each person shakes hands with every other person exactly once, how many total handshakes happen?
@phi @amistre64 @AkashdeepDeb @hartnn @ranga

Answer 1

36

Answer 2

There are no choices of 36.

Answer 3

what are the choices?

Answer 4

From 6 people you have TO SELECT 2 poeple for a hand shake.

So the solution would be 6C2 = 6 * 5 / 2 * 1 = 15 ways! :D

Answer 5

you wouldn't know because they shake hands with every other person?

Answer 6

@AkashdeepDeb Can you give me a more easier way to solve this problem?

Answer 7

Do you know combinations? :)

Answer 8

No

Answer 9

Can u explain it to me :)

Answer 10

Sure, but first, Do you know factorials?

Answer 11

No

Answer 12

I think then it'll take you some time and wouldn't be possible to teach on OS.

Answer 13

http://www.coolmath4kids.com/math_puzzles/h1-handshakes_sol5.html

Answer 14

one way is say: one person will shake hands with 5 other people
there are 6 people so 5*6 = 30
but, we are double counting (A with B is the same as B with A)
so 30/2 = 15 handshakes.

Answer 15

http://mathforum.org/library/drmath/sets/select/dm_handshakes.html

Answer 16

but 6 choose 2 is a nicer way to solve this problem

Answer 17

@phi 's explanation Is TOTALLY CORRECT.

That IS in fact, in short, the function of combinations in probability and in the fundamental principal of counting.

Answer 18

Thanks @AkashdeepDeb and @phi. I learned something new today :)

Answer 19

:)

Answer 20

one more way to look at it is,
the first person will shake hands with 5 other person, so \(\huge5\)
the 2nd person already shaked his hand with the 1st person, remaining shakes = \(\huge4\)
continuing this way, we get
5+4+3+2+1 = 15 ways

infact there a formula dedicated to this handshake problem, which directly comes from sum of 'n' natural number formula,
for n ppl, total handshakes will be
\(\large \dfrac{n(n-1)}{2}\)