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

Guest Apr 30, 2018

#1**0 **

wouldnt it just be 8 because if they shake each others hand and there is 10 people but excluding john and his wife

Guest May 1, 2018

#2**+1 **

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 |

Melody May 1, 2018