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Consider the set of all functions from $\{1, 2, . . . ,m\}$ to $\{1, 2, . . . , n\}$,where $n > m$. If a function is chosen from this set at random, the probability that it will be strictly increasing is

  1. $\binom{n}{m}/n^m\\$
  2. $\binom{n}{m}/m^n\\$
  3. $\binom{m+n-1}{m-1}/n^m\\$
  4. $\binom{m+n-1}{m}/m^n$

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Total no. of functions is $n^m$. [For every value from ${1,2....,n}$ there are m possibilities].

We choose an m-element subset of n, number of such subsets $=\binom{n}{m}$

Since it is an increasing function, we can order the elements in only one way, i.e in ascending order.

Therefore , probability is $\frac{\binom{n}{m}}{n^m}$
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