Question

[SOLVED] Here's a neat little combinatorics/graph theory question.

Mr. Smith and his wife invited four other couples for a party. When everyone arrived, some of the people in the room shook hands with some of the others. Of course, nobody shook hands with their spouse and nobody shook hands with the same person twice.

After that, Mr. Smith asked everyone how many times they shook someone’s hand. He received different answers from everybody.

How many times did Mrs. Smith shake someone’s hand?

PS. This took me about 5 minutes to solve when I was presented with it. Can you beat me?

Answer 1

8 people.

Answer 2

I mean, if four other couples were invited, then 4*2 = 8

Answer 3

it's a little more complicated than that. Remember that everyone shook hands a different number of times (with Mr. Smith as the exception).

Answer 4

Yes, I excluded Mr. Smith.

Answer 5

Btw, the answer isn't 8.

Answer 6

Barely legible hint in white LaTeX below.\[ \color{white}{\text{what is the maximum/minimum number of times} \text{someone could shake hands?} }\]

Answer 7

0? Hmm

Total of 10 people and Mr. Smith shook hand with everyone right? 8 is the maximum number of handshakes? and you're asking Nine people. So, 0 for Mrs. Smith?

Answer 8

It isn't 0 either. Perhaps draw a picture, and start drawing handshakes with some random person. Also, remember that he asked 9 people how many handshakes they did. What is the maximum number of handshakes a single person could have?

Answer 9

isn't it 0 and 8 as max/min
you can't shake your own hand nor your spouse, which leaves 8 people

Answer 10

he received a different answer from everyone...0 thru 8 gives 9 different answers
right

Answer 11

That's exactly the conclusion to make. Good work so far.

Answer 12

Perhaps another leading question. How many people shook hands 8 times? How many 0 times?

Answer 13

i believe its impossible for 2 people to have shook hands 0 or 8 times
thus mr. smith shook hands somewhere between 1 and 7
following that logic, i will say 4 times is answer

Answer 14

Correct. Basically, he didn't shake hands with himself or his wife. So that leaves a maximum of 8. One person shook hands 0 times, so max of 7. One person shook hands 1 time, but it had to have been with the person who shook hands 8 times. So we have a max of 6.

Continue this way, and you get a max of 4. At this point, if you draw it out, 5 people have shaken hands 0, 1, 2, 3, 4 times, and 4 people have shaken hands 4, 5, 6, 7 times. To fix this, Mr. Smith has to shake hands with the last 4 people.

So 4 shakes.

Answer 15

Can you figure out how many times his wife shook hands?

Answer 16

well its not 0 for sure...kinda stuck, i think its possible she could be the one to have shaken 8 times ?

Answer 17

Suppose she did shake 8 times. Who would have shaken 0 times?

Answer 18

oh duh, haha ok so not 8

Answer 19

wait i see the pattern
for this to work each couple must complement each other so their sum will be 8 handshakes
0,8
1,7
2,6
3,5
4,4

Mrs. smith shook 4 times as well

Answer 20

Correct again!

Answer 21

I was actually going to ask "How many times did each couple shake hands next" :P

Looks like you already solved that.

Answer 22

haha :P
these are fun...takes me awhile to visualize the problem

Answer 23

The way I see it for these problems, is to draw it, and to see what happens at extremes. I.e. seeing what would happen if Mr. Smith shook hands 8 times/0 times. Then try to extrapolate.