I'm sure there's gotta be a better way than this, but here's how I did it:

5 cards can be selected from 10 cards 10!/(5!*(10-5)!) or 252 ways.

Then I counted out all iterations where the average equals the mean:

- There are 6 consecutive instances (1,2,3,4,5 --> 6,7,8,9,10)
- There are 4 instances with a gap in the 1st and 4th slots (1,3,4,5,7 --> 4,6,7,8,10)
- There are 4 instances with a gap in the 2nd and 3rd slots (1,2,4,6,7 --> 4,5,7,9,10)
- There are 2 instances with 2 gaps in the 1st and 4th slots (1,4,5,6,9 --> 2,5,6,7,10)
- There are 2 instances with 2 gaps in the 2nd and 3rd slots (1,2,5,8,9 --> 2,3,6,9,10)
- There are 2 instances with a gap in all 4 slots (1,3,5,7,9 --> 2,4,6,8,10)

This equals 20 total instances where the median equals the mean.

So there are 232 instances where the median does not equal the mean. 1/2 the time the mean > median and 1/2 the time the median > mean, so the answer is 116, or B.

I'd like to see a better way to do it if anyone has it though. I didn't know how else to find out the situations where the mean = median.

knight247 wrote:Five cards are to be selected from 10 cards numbered 1 to 10. In how many ways can the cards be selected so that the average of the five numbers selected is greater than their median?

(A) 111
(B) 116
(C) 222
(D) 232
(E) 252

Don't have an OA on this. Detailed explanations would be highly appreciated.

The number of ways to choose 5 cards from 10 choices = 10C5 = 252.

We need to subtract from 252 the number of combinations in which the average = the median:

Median and average of 3:
{1,2,3,4,5}
Number of options = 1.

Median and average of 4:
{1,2,4,5,8} {1,2,4,6,7} {1,3,4,5,7} {2,3,4,5,6}
Number of options = 4.

Median and average of 5:
{1,2,5,7,10} {1,2,5,8,9} {1,3,5,6,10} {1,3,5,7,9} {1,4,5,6,9} {1,4,5,7,8} {2,3,5,6,9} {2,3,5,7,8} {2,4,5,6,8} {3,4,5,6,7}
Number of options = 10.

Median and average of 6:
{1,4,6,9,10} {1,5,6,8,10} {2,3,6,9,10} {2,4,6,8,10} {2,5,6,7,10} {2,5,6,8,9} {3,4,6,7,10} {3,4,6,8,9} {3,5,6,7,9} {4,5,6,7,8}
Number of options = 10.

Median and average of 7:
{3,6,7,9,10} {4,5,7,9,10} {4,6,7,8,10} {5,6,7,8,9}
Number of options = 4.

Median and average of 8:
{6,7,8,9,10}
Number of options = 1.

Total combinations in which the average = the median: 1+4+10+10+4+1 = 30.

Thus, we have 252-30 = 222 combinations in which the average ≠ median.

In half of these combinations, the average will be less than the median; in the other half, the average will be greater than the median.

Thus, the number of combinations in which the average is greater than the median: 222/2 = 111.

The correct answer is A.

Please note that this question would NOT be asked on the GMAT. WAY too time-consuming.

Ahhh, thanks for the clarification GMATGuru. I was fixated on symmetric distributions around the median, so I missed out on the other 10 instances that were asymmetric.