Question

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Answer 1

it takes 2 people to shake hands

Answer 2

I understand the concept of handshakes but I am not getting the correct meaning. Yes I know 2 people = 1 pair

Answer 3

E.g. if 3 people = 3 pairs possible

Answer 4

look at points and how they can connect. start simple and then build.

Answer 5

I am not understanding the question.

Answer 6

the max number of handshakes, not including repeats for any 1 person is n-1 where n = total number of people. this also assumes one can't (or doesn't) shake hands with him/her self.

Answer 7

the total number of handshakes must alway be even.

Answer 8

Correct, suppose 3 people ABC, A can shake hand with B & C which is n-1, am I right?

Answer 9

or, the total number of people who shook hands must always be even

Answer 10

yes

Answer 11

AB, AC , BC, For person A= AB, AC, for person B= BA, BC, for person C= BC, AC

Answer 12

correct?

Answer 13

yeah

Answer 14

Thanks.

Answer 15

that doesn't prove your problem, though

Answer 16

I knew the answer, just could not understand the question.

Answer 17

hmm. so how would I prove it?

Answer 18

seems like its a combination problem.

Answer 19

assume there are n people at the party.

assume that all shook a different number of hands. (which means there is not at least one pair of people who shook the same number of hands).

show that this can't happen for n =2, 3, ...

Answer 20

can you show with 3 people please?

Answer 21

ABC = AB & C; A shook once, B shook once, C did not shake. but B and A shook one time each O.o

Answer 22

ah its pair.

Answer 23

let's do n = 2 first.

suppose there are 2 people at a party. since a person can't shake hands with themselves (or at least we don't consider that a handshake), then we can't have person A with 0 handshakes and person B with 1 handshake.

the total number of possible handshakes = {0, 2} which is even.

but 0+1 = 1 is odd.

Answer 24

wait AB = 2 handshake?

Answer 25

for 3 people, we have a total of {0, 2, 4, 6} handshakes. the max any one person can have is 2.

therefore, for all of them to have a different number of handshakes we'd have to have

0, 1, 2 => a total of 3 which is odd

Answer 26

See its the definition that's causing trouble, so if A&B handshake its 1 pair but 2 handshakes, 1 for each?

Answer 27

if there are and even number of people (like when n = 4), then we get things that can't happen.

for example, we would need them all to have a different number of handshakes.

so let's say A would have to have 0 handshakes, B would have 1 handshake, C would have 2 handshakes and D would have 3 handshakes. 3 would be the mak anyone could have.

but we can have 0 and 3 as a choice, because in order for D to have 3, D would have to shake everyone's hand (again, assuming no repeats). Thus, A couldn't be 0!

Answer 28

"max" not "mak"

Answer 29

i guess the thing is you can't have both someone having the max number of handshakes and another having the minumum, regardless of what n is.

Answer 30

have you learned the pigeon hole principle?

Answer 31

ye, if some thing does not fit in, it has to be forced.

Answer 32

eg 6 hole, 7 pigeons, 1 need to be forced, right?

Answer 33

right, so you could apply that here.

Answer 34

your choices for the number of hands shook is 0, 1, 2, 3, ..., n-2 or 1, 2, 3, ..., n-1

you can't have 0 AND n-1 in the same group of choices. it simply isn't possible!

now each of those groups contains n-1 elements and you can't assign n elements to n-1 elements without at least 1 repeat.

make sense?

Answer 35

I am getting the idea now, so for example if 2 people, its either 0 pair (none shook hands) and 1 pair (A & B shook hands) but if A shook hand then so did B and if both did not shake hand then its 0 for both right?

Answer 36

AB = 1 pair but 1 handshake for each A & B. So A shook once and B shook once.

Answer 37

not quite,

your choices are 0 handshakes or 1 handshake.

let's say 0 hand shakes.

then you have to assign each of the 2 people to that choice.

likewise, if you say 1 handshake, then you have to assign both to that choice.

it isn't possible that on person shakes everyone's hand at the party and for another to not shake anyone's hand. because if one person shook everyone's hand, that means everyone has shaken that person's hand and no one can claim they haven't shook hands with anyone.

Answer 38

that's what I am saying, if 0 handshakes, 0 for A and 0 for B , if 1 handshake, 1 for A and 1 for B

Answer 39

for 3 people; ABC, If A shakes hand with B and C but B and C do not shake hands then its 2 for A, 1, for B, 1 for C. but if B and C shake hands then its 2 for A, 2 for B, 2 for C. If none shake hand then its 0 for all of them.

Answer 40

think of it as how you assign an element of one set to another.

for2 people, we have 2 sets of handshakes {0} and {1}, we can't have {0,1} as a possible set for the reason i explained above.

now we have 2 people, say {A, B}

we can't assign with overlap!