Question
Show that there must be at least 90 ways to choose six integers from 1 to 15 so that all choices have the same sum.
Answers
Steve
There are 6C15 = 5005 ways to choose 6 numbers
The minimum sum is 1+2+2+4+5+6 = 21
the maximum sum is 10+11+12+13+14+15 = 75
Thus there are only 55 different sums possible
5005/55 = 91
So, at least one of the sums must appear at least 91 times.
The minimum sum is 1+2+2+4+5+6 = 21
the maximum sum is 10+11+12+13+14+15 = 75
Thus there are only 55 different sums possible
5005/55 = 91
So, at least one of the sums must appear at least 91 times.
Monika v
Thank u for u r answer giving and continue this work . I'm asking for again and again questions u respond it.