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The original statement was completely wrong. Here's the actual question. How many ways are there of picking n numbers from the set {1,...,N} where exactly m of the numbers picked are less than or equal to M. (Picking with replacement).

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    If $1 \leq M < n$, there are infinitely many ways: choose the number $1$ exactly $m$ times, and then choose the number $n$ as many times as you like. When $n > 2$ there are additional ways not counted above, but they don't change the number of ways (which are already countably infinite). Perhaps there is another parameter missing from the question, such as how many numbers you are supposed to pick? – 2017-01-30
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    Yes I've updated the question now. In fact I messed up the whole question when typing it out. – 2017-01-30

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Any number has a $(\frac MN)$ probability of being less than or equal to $M.$

Then there are exactly $(n-m)$ that are greater than $M.$ And that has as $(1-\frac MN)$ probability.

And finally there are ${n\choose m}$ ways do distribute the selections.

${n\choose m} (\frac MN)^m (1-\frac MN)^{(n-m)}$

Moving out of probabilty space into "how many ways space" Then multiply the above by how many ways can you pick $n$ numbers from the set $\{1,2,\cdots,N\} = N^n$

${n\choose m} M^m (N-M)^{(n-m)}$

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    How are there $(n-m)$ numbers greater than $M$? Surely it should be $(N-M)$. – 2017-01-30
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    You have selected $n$ numbers from the set, $m$ of which are $\le M.$ $n-m$ of your selection are $< M \cdots N-M$ members of the set are $\le M$ – 2017-01-30
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    What about the number of ways of we were repeating the experiment but without replacement. I got the number of ways as $\frac{M!}{(M-m)!} \cdot \frac{(N-m)!}{(N-n)!} \cdot \begin{pmatrix} n \\ m \end{pmatrix} $. – 2017-01-31
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    without replacement.. I think it would be ${n\choose m} {M\choose m} {N-M\choose n-m}$ – 2017-01-31
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Hint: since you're choosing with replacement, the distribution of each draw is the same and independent from the last. The chance that a single draw is less than or equal to $M$ is $M/n$. Alternatively, there are exactly $M$ numbers that qualify as $\leq M$ on each draw. Now, have you heard of the binomial distribution? Otherwise, how many ways are there to pick exactly $m$ numbers less than or equal to $M$?

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    Even this simple question I'm not really too sure. Is it $M^m$? – 2017-01-30
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Here's the actual question. How many ways are there of picking n numbers from the set {1,...,N} where exactly m of the numbers picked are less than or equal to M. (Picking with replacement).

Note : Assuming integers, that $0\leq m\leq M\leq N$, $(n-m)\leq (N-M)$ (why?), and that order is irrelevant.

You wish to count distinct selections of $m$ from $M$ and $(n-m)$ from $(N-M)$, when selecting any $n$ from all $N$ distinct items without replacement.

The binomial coefficient, $\binom K k$, also written as ${^K\!\mathrm C_k}$, counts ways to select $k$ from $K$ distinct items without replacement where order does not matter.   So ...

$$\binom K k = \dfrac{K!}{k!~(K-k)!}$$