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Mock AIME 4 2006-2007 Problems/Problem 8

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Problem

The number of increasing sequences of positive integers a_1 \le a_2 \le a_3 \le \cdots \le a_{10} \le 2007 such that a_i-i is even for 1\le i \le 10 can be expressed as {m \choose n} for some positive integers m > n. Compute the remainder when m is divided by 1000.

Solution

The numbers a_i - i are ten not-necessarily distinct even elements of the set \{0, 1, 2, \ldots, 1997\}. Moreover, given ten not-necessarily distinct elements of \{0, 1, 2, \ldots, 1997\}, we can reconstruct the list a_1, a_2, \ldots, a_{10} in exactly one way, by adding 1 to the smallest, then adding 2 to the second-smallest (which might actually be equal to the smallest), and so on.

Thus, the answer is the same as the number of ways to choose 10 elements with replacement from the set \{0, 2, 4, \ldots, 1996\}, which has 999 elements. This is a classic problem of combinatorics; in general, there are {m + n - 1 \choose m} ways to choose m things from a set of n with replacement. In our case, this gives the value of {999 + 10 - 1 \choose 10} = {1008 \choose 10}, so the answer is 008.


Note that in fact, the answer is not unique because a many numbers can be represented as a binomial coefficient in more than one way. For instance, another such expression for the number of such sequences is {{1008 \choose 10} \choose 1}. It so happens that these (with their corresponding expressions {m \choose m - n}) are the only four times this number appears in Pascal's Triangle.

See also

Mock AIME 4 2006-2007 (Problems, Source)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Want to learn how to tackle those tough AMC/AIME/Olympiad algebra problems? Check out Art of Problem Solving's Intermediate Algebra by Richard Rusczyk and Mathew Crawford. Over 1600 problems!
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