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 17:02, 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.
