Question

1. How many handshakes?

Answer 1

Ten total.

Answer 2

There are twenty ordered pairings, but the order (who is the "lead" handshaker in ea pair) doesn't matter.

Answer 3

I think the way your instructor wants you to see it is this:

Answer 4

First person in room shakes hands with no one.
Second shakes hands with one.
Third shakes hands with two.
Fourth with three.
Fifth with four.
0+1+2+3+4 = 10.

Answer 5

welcome.

Answer 6

you can use binomial number
\[\left(\begin{matrix}5 \\ 2\end{matrix}\right)\]
i read it as (from 5 chose 2)
\[\left(\begin{matrix}5 \\ 2\end{matrix}\right)=5!/(2!)(3!)=5*4/2=2*5=10\]

this is always used to calculate combinations where it doesn't matter the combination (2-1=1-2)

Answer 7

I look at it like this: 5 people in the room. Each person shakes the hands of 4 other people. 5 x 4 = 20 handshakes total / 2 (because person A shaking person B's hand is the same as person B shaking person A's hand) = 10 total handshakes.

Answer 8

Imagine a "handshake" booth. There are two spots for people, the two people shaking hands. How many ways can you arrange 5 people in that room where the order of those people don't matter?

Well, we can put 5 people in the first spot, but for our second spot we can only put 4 because one of them is already in the room. That gives us 5 * 4 which = 20.

Now we must think about what it means for "order not to matter". Say A and B are in the booth. That's the same as B and A. How many ways can 2 people be arranged? Well, just two ways. So, we divide by 2.

Answer is 20/2 = 10.

This is a combinations problem, and the formula for combinations is: (n!)/((n - k)! * k!) where n is the total number of things to choose from and k is the total number of spots. 5 and 2 in this case, respectively.

Answer 9

you can also think physically.
get 5 people. the first one is going to handshake everyone, so here is 4 handshakes.
the second person is just going to handshake 3 people because he already handshaked the first guy. handshaking again would be silly
the third person has just 2 other left to handshake, the fourth, just one. and the last has already handshaked everyone else and one handshaking oneself is also silly.
so you have 4+3+2+1=10 handshakes...

this exp is just for u to get the ideia of what the binomial number mean. in more complex problems just use \[\left(\begin{matrix}a \\ b\end{matrix}\right)\]