Difference between revisions of "2011 AIME II Problems/Problem 6"

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Summing, 34 + 22 + 13 + 7 + 3 + 1 = 080, answer
 
Summing, 34 + 22 + 13 + 7 + 3 + 1 = 080, answer
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'''Solution 2'''
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There are "10 choose 4" total ways to choose a, b, c, and d(210).
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Do casework to find that there are 50 cases where a+d=b+c.  The number of cases where a+d}c is equal to the number of cases where a+d{c.  Therefore the answer is (210-50)/2='''80'''
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I apologize for the lack of explanation; I don't have enough time.

Revision as of 18:08, 31 March 2011

Problem:

Define an ordered quadruple (a, b, c, d) as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many ordered quadruples are there?


Solution:

There is probably some really complicated formula for this, but as I didnt know it and had 3 hours to "do my best", I listed all possible combinations out. The answer is 80. You can do casework with the value of a. If a = 1, we can do casework on b. If b = 2, we can do casework on c. if c = 3, d = 5,6,7,8,9, or 10. That's 6. If c = 4, d can be 6,7,8,9,10. That's 5. Continuing this pattern, we get 21 options is b = 2. if b=3 and c = 4, d = 7, 8, 9, or 10, c= 5 gets d = 8, 9, 10 and by quick inspection we get 10 options with b = 3. With b = 4, c = 5 we get d = 9,10, c= 6 we get d = 10, that's 3, with b = 5 there are no solutions.

So if a = 1 there are 21+10+3 = 34 options

If a = 2, b = 3, c= 4, then d = 6,7,8,9,10 so 15 options with b = 3 (same pattern as above). if b = 4, c=5 gives d = 8, 9, 10 so 6 options. if b = 5, c=6 gives just one option, d = 10, and if b=6, there are no options.

So if a = 2 there are 15+6+1 = 22 options

If a = 3, b = 4, c = 5, then d = 7, 8, 9, 10 so 10 options (keeping with the same pattern). if b=5 and c = 6, d = 9,10 so 3 options. If b = 6 and c = 7, there are no options.

So if a = 3 there are 10+3 = 13 options

If a=4, b=5, c= 6, then d = 8, 9, 10 so 6 options. If b = 6, c= 7 there is just one option

So if a = 4 there are 6+1 = 7 options.

If a = 5, b = 6, c = 7, d can be 9, 10 so 3 options. If b = 7, c = 8 there are no options

So if a = 5 there are 3 options.

Finally, If a = 6, b = 7, c =8, there is just one option, d = 10

So if a = 6 there is just 1 option

Summing, 34 + 22 + 13 + 7 + 3 + 1 = 080, answer

Solution 2

There are "10 choose 4" total ways to choose a, b, c, and d(210).

Do casework to find that there are 50 cases where a+d=b+c. The number of cases where a+d}c is equal to the number of cases where a+d{c. Therefore the answer is (210-50)/2=80

I apologize for the lack of explanation; I don't have enough time.

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