Question 1162881: Three distinct integers are chosen from the set {1, 2, 3, . . . , 19, 20} and arranged from smallest to largest. What is the number of ways of choosing these three integers so that the middle integer of the three is greater than or equal to the average of the largest and the smallest?
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
615
I'll outline a couple of strategies for finding this answer and let you do the work to fill in the details.
First strategy....
Find the number of sequences of three of the numbers that satisfy the condition if the smallest of the numbers is 1.
Find the number of sequences of three of the numbers that satisfy the condition if the smallest of the numbers is 2.
Find the number of sequences of three of the numbers that satisfy the condition if the smallest of the numbers is 3.
...
Find the number of sequences of three of the numbers that satisfy the condition if the smallest of the numbers is 17.
Find the number of sequences of three of the numbers that satisfy the condition if the smallest of the numbers is 18.
Add up all those different numbers to get the answer of 615.
Don't work all the cases out completely by hand; instead, look for a pattern in the numbers for the cases with increasing smallest numbers, and use that pattern to reach the final answer.
Second strategy....
The three numbers chosen will have the middle number either (a) less than, or (b) equal to, or (c) greater than the average of the first and last.
By symmetry, the number of combinations with the middle number less than the average of the first and last will be the same as the number of combinations with the middle number greater than the average of the first and last.
So...
(1) determine the number of combinations of three numbers you can choose out of 20: 20C3 = 1140
(2) determine, using a case-by case analysis similar to the above, the number of combinations in which the middle number is EQUAL to the average of the first and last. You should get a total of 90.
(3) So the number of sequences in which that is NOT true is 1140-90 = 1050. Then the number of sequences with the middle number greater than the average of the first and last is 1050/2 = 525.
(4) And then the number of sequences with the middle number greater than or equal to the average of the first and last is 525+90 = 615.
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