THOMAS GOES TO THE BIRTHDAY PARTY OF HIS FRIEND & FOUND THAT THERE ARE 3 GUESTS WHO HAVE B'DAY ON THE SAME DAY OF THE WEEK & IN THE SAME MONTH OF THE YEAR.HE ALSO FOUND THAT ALL THE GUESTS WERE BORN IN THE FIRST SIX MONTHS OF THE YEAR.FIND THE LEAST NO. OF PEOPLE INCLUDING THOMAS AT PARTY. . . . ANSWER IS '85' HOW TO GET ANSWER ? PLEASE HELP ME!!!!!!!!!!

## Comments

Sort by:

TopNewestThere are 6 different months and 7 different days of the week, so there are 42 different possible combinations of a month and a week. For there to be 3 of the same month and same week, by Pigeonhole principle there must be at least \(42 \times (3-1)+1=85\) people. – Joel Tan · 3 years, 4 months ago

Log in to reply

– Harsh Shrivastava · 3 years, 4 months ago

please what is piegon hole principle ???????????????????????????Log in to reply

Or generally, for your question, this version is better: if there are x pigeons and y pigeonholes, one of the pigeonholes must contain at least n pigeons where n is the smallest integer larger than x/y.

It can be proved by contradiction: suppose that all pigeonholes contain less than n pigeons. Then each pigeonhole contains less than \(\frac{x}{y}\) pigeons. However, since there are y pigeonholes the total number of pigeons is less than x, contradicting the fact that there are x pigeons.

So in your question, there are 42 different 'pigeonholes' as mentioned above. Also, by the above, the smallest integer larger than \(\frac{number of people}{42}\) is 3. So the fraction must be larger than 2 so the number of people is greater than 84, so 85 is minimum. – Joel Tan · 3 years, 4 months ago

Log in to reply

– Harsh Shrivastava · 3 years, 4 months ago

thanks a lot !Log in to reply