anonymous
  • anonymous
1. How many handshakes?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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rulnick
  • rulnick
Ten total.
rulnick
  • rulnick
There are twenty ordered pairings, but the order (who is the "lead" handshaker in ea pair) doesn't matter.
rulnick
  • rulnick
I think the way your instructor wants you to see it is this:

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rulnick
  • rulnick
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.
rulnick
  • rulnick
welcome.
anonymous
  • anonymous
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)
anonymous
  • anonymous
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.
anonymous
  • anonymous
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.
anonymous
  • anonymous
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 \[\left(\begin{matrix}a \\ b\end{matrix}\right)\]

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