For some of those picks of 5 cards, the mean will equal the median. The rest will be equally split between having a higher mean or a higher median. So we can solve if we know all the way in which mean and median will be the same.
Great start.
Median will be between 3 and 8 inclusive.
for 3 and 8 we only have one way to choose the rest of the 4 numbers
1 2 3 4 5
6 7 8 9 10
Between 3 and 8
4, 5, 6, and 7
For example FOR Median = Average = 4
2 3 4 5 6
We will have only (n-1) number on the left to play with 2 numbers and every time you make a change you have to make another change on the other side and you can choose 2 of them
For 4 = REST of the sum will be 16 ( you can use 1,2 or 3) C(3:2) = 3
For 5= REST of the sum will be 20 (you can use 1, 2, 3 or 4) C(4:2) = 6
For 6= REST of the sum will be 24 ( you can use 7,8,9 or 10 ) C(4:2) = 6
For 7= REST of the sum will be 28 ( you can use 8,9 or 10 ) C(3:2)=3
So AV=MEDIAN= 1+3+6+6+3+1 = 20 ways ( If I am not mistaken)
C(10:5)= 252
252 - 20 = 232 way for > Median and < Median
So answer should be 232/2 = 116
:roll:
I realized that I have missed couple of numbers!
median=3: only one way: 12345
median=4: four ways: 12467, 12458, 13457, 23456
median=5: ten ways: 12589, 1257T, 13579, 1356T, 14569, 14578, 23578, 23569, 24568, 34567
(I listed the numbers lower than the median systematically, then looked for pairs of numbers higher than the median that added to the right sum)
median=6 ten ways (symmetric with median=5)
median=7,four ways (symmetric with median=4),
median=8, 1 way 6789T
Total ways = 1+4+10+10+4+1 = 30
This makes the correct answer: 111
252-30 =222
and 222/2 = 111
