Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 8"
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− | 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 <math>{{1008 \choose 10} \choose 1}</math>. It | + | 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 <math>{{1008 \choose 10} \choose 1}</math>. It so happens that these (with their corresponding expressions <math>{m \choose m - n}</math>) are the only four times this number appears in [[Pascal's Triangle]]. |
Revision as of 18:02, 10 February 2007
Problem
The number of increasing sequences of positive integers such that
is even for
can be expressed as
for some positive integers
. Compute the remainder when
is divided by 1000.
Solution
The numbers are ten not-necessarily distinct even elements of the set
. Moreover, given ten not-necessarily distinct elements of
, we can reconstruct the list
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 , which has 999 elements. This is a classic problem of combinatorics; in general, there are
ways to choose
things from a set of
with replacement. In our case, this gives the value of
, so the answer is
.
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 . It so happens that these (with their corresponding expressions
) are the only four times this number appears in Pascal's Triangle.