nCk=n!/k!(n-k)! used in 'handshake' problem

[kwoodward]

One of my students in 8th grade brought me this problem. I am teaching an all level math class but don't recall this equation from back in my day doing math. The problem is this: After a band parent meeting, each of the ten members shook hands with each other once. How many hand shakes were there in all? The formula nCk=n!/k!(n-k)! is supposed to be utilized.

HELP! Please solve entire problem if you can so I can study! :roll:

Thank you! KW

 

[tkhunny]

2 people - 1 handshake
3 people - 3 handshakes - 2 more
4 people - 6 handshakes - 3 more
5 people = 10 handshakes - 4 more
6 people = 15 handshakes - 5 more

I see a pattern.

n = 2
k = 2
\(\displaystyle \frac{2!}{2!0!}\;=\;1\)

n = 3
k = 2
\(\displaystyle \frac{3!}{2!1!}\;=\;3\)

n = 4
k = 2
\(\displaystyle \frac{4!}{2!2!}\;=\;6\)

etc.

Where are you struggling?

 

[Denis]

Hmmm...can't see why that formula is suggested;
much simpler/easier one:
n(n - 1) / 2 where n is the number of persons