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Find the number of sequences $(a_1,\cdots,a_{100})$ where all $a_i$'s are even numbers which forms $0\leq a_1 \leq \cdots \leq a_{100} \leq 200$.

I know that the number of strictly incremental sequences of length $K$ with elements from $\{1, 2,\cdots,N\}$ is ${N}\choose{K}$. However, here the sequence is not strictly increasing sequence, hence the same answer should not be valid. The answer to my question is ${200}\choose{100}$, which feels weird to me because there are a lot more options for us to pick the sequences because it is not strictly increasing, however this is the same answer as the ${N}\choose{K}$ from the strictly increasing sequences.

Why the solution for both problems are the same?

ofirsasoni
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    "which feels weird to me because there are a lot more options for us to pick the sequences because it is not strictly increasing", but in this question you are only picking from the even numbers, and there are only $101$ even numbers in $0\le a \le 200$. – peterwhy Mar 12 '24 at 18:28
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    Just because things look the same, doesn't necessarily mean that they are the same. And if they are numerically equal, sometimes there could be a bijection between the two of them. – Calvin Lin Mar 12 '24 at 19:17
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    Divide each term by $2$, $\left(\frac {a_i}2\right)_{i=1}^{100}$ is a non-decreasing sequence of (even or odd) integers satisfying

    $$0 \le \frac{a_1}2 \le \frac{a_2}2 \le\ldots\le \frac{a_{100}}2\le 100.$$

    Increase each term by its index $i$, $\left(\frac {a_i}2 + i\right)_{i=1}^{100}$ is a strictly incremental sequence of integers satisfying

    $$0<\frac{a_1}2+1<\frac{a_2}2+2<\ldots<\frac{a_{100}}2+100 < 201.$$

    This shows one way of the bijection.

     – peterwhy Mar 12 '24 at 19:18

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