# Pigeonhole Principle – 3

## 10 thoughts on “Pigeonhole Principle – 3”

1. In the problem where we are asked to find the probability of handshake by two,,please elaborate the conclusion.I din’t understand the last few lines after 99 handshakes.

• If a person shakes hands with all 99 others, then there cannot be anyone with zero handshakes, right? So basically the possible values for “number of handshakes” can range from 1 to 99 or from 0 to 98. Either way a max of 99 distinct values.

regards
J

2. Ziddi Armaan |

@ J : Can u explain why we multiply 366*5 ??

3. Girish |

yup sir i too had a doubt regarding 6 people sharing same bday… i got the answer as 371 : 366 plus 5

4. 366*5 is the “worst case scenario not satisfying the condition”. We could have these many people, and still have not more than 5 people with a birthday on any given day.

For the handshakes: there are 100 people and 99 possible values so not everybody can have different values.

regards
J

5. Parveen |

J, could you please explain the intent of the last question? I din’t quite get it. Thank you!

• No intent! Unless you count “confusing the student” as the examiners’ intent ðŸ™‚ It is just a typical question which can be solved using the pigeonhole logic. The idea being again that since there are 26 things to be put into 25 regions, at least one the regions must contain more than one thing.

regards
J

6. In the problem where we are asked to find the probability of handshakes,could you please elaborate the conclusion? I didnâ€™t understand the last few lines after 99 handshakes.

7. Rohit Gupta |

I got everything, from region one to the handshakes and hair, but birthdays still went overhead. Can you please explain that again, Sir? 366 birthdays. Can’t there be simply 5 sharing birthday on any specific day among these 366?

• There can be. For that 5 is enough. But we are asked to ensure that there will be. Certainty, not possibility, is demanded.

regards
J