Five married couples sit down to
play poker (10 people). Before
they play, each person shakes the
hand of everyone at the table that
they have not met. After they
shake hands, John asks everyone
else how many hands they shook.
The nine other players respond:
8, 7, 6, 5, 4, 3, 2, 1, and 0.
How many hands did
John shake? How
many hands did
John’s wife shake?
I got this from agmath.com
wouldnt it just be 8 because if they shake each others hand and there is 10 people but excluding john and his wife
+118609
8, 7, 6, 5, 4, 3, 2, 1, and 0.
1) No one will shake hands with themself or their spouse so I can put a X in each of those pairs
2) Now, let AH be the one that shakes 8 hands
3) Now all have shaken AH hand axcept for his wife so his wife must be the only one to shake no hands
4) Let BH be the one to shake 7 hands
5) now all have shaken 2 hands already except for BW so she must be the one to shake just one hand
6) Let CH be the one to shake 6 hands
7) Now they have all shaken 3 hands except for CW so she must be the one to shake only 2 hands.
8)Let DH be the one to shake 5 hands
9) which means that D wife shakes 3 hands
Which now leaves E husband and E wife to shake 4 hands each.
That is, John and his wife shake 4 hands each :)
| AH | AW(3) | BH(4) | BW(5) | CH(6) | CW(7) | DH(8) | DW(9) | EH | EW | count | |
| AH | X | X | y | y | y | y | y | y | y | y | 8 |
| AW(3) | X | X | X | X | X | X | X | X | X | X | 0 |
| BH(4) | y | X | X | X | y | y | y | y | y | y | 7 |
| BW(5) | y | X | X | X | X | X | X | X | X | X | 1 |
| CH(6) | y | X | y | X | X | X | Y | Y | Y | Y | 6 |
| CW(7) | y | X | y | X | X | X | x | x | x | x | 2 |
| DH(8) | y | X | y | X | Y | x | X | X | y | y | 5 |
| DW(9) | y | X | y | X | Y | x | X | X | x | x | 3 |
| EH | y | X | y | X | Y | x | y | x | X | X | 4 |
| EW | y | X | y | X | Y | x | y | x | X | X | 4 |
| count | 8 | 0 | 7 | 1 | 6 | 2 | 5 | 3 | 4 | 4 |
