some people go to a party there is a total of 5 handshakes. How many people are at the party?

Question

kainui · Answer

"Party"
"5 handshakes"

kainui · Answer

Try drawing out a picture with dots, and lines between the dots represent handshakes.

anonymous · Answer

Dots representing handshakes? You just blew my mind.

kainui · Answer

no no no, lines are the handshakes. The dots are the people.

anonymous · Answer

dats wut i ment

anonymous · Answer

sorry ma english is not so gud

kainui · Answer

I don't know how looking at how many considerate people are attending a party has anything to do with the total number of people attending.

anonymous · Answer

It doesn't lol, I think op made this question up or is missing something.

mathmale · Answer

Here's my point of view:  Only two people are involved in any one handshake.  Let's say that there are n people at the party; we don't know what n is yet.  All we know is that five separate handshakes take place.

Think "combinations."  $$_{n}C _{x}$$represents the number of combinations possible from a set of n, taken x at a time.

It happens that this is equal to $$\frac{ n! }{ x!(n-x)! }$$

So here, with n unknown, but x known to be 2, and the total number of combos known to be 5, we write the equation$$\frac{ n! }{ 2!(n-2)! }=5$$ and solve it for the single unknown, n.  

Anyone?

mathmale · Answer

It occurs to me that n has to be at least 2, because of that (n-2)! term in the denominator.

mathmale · Answer

Note:  I've tried n = 2, 3, 4 and 5, but so far have not found that any of these possible values for n "works."  Comments?  Suggestions?

triciaal · Answer

4 with one missing handshake