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This is a beautiful problem from ISI MStat 2018 PSA problem 13 based on probability of functions. We provide sequential hints so that you can try .

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

- \( {n \choose m} / m^{n} \)
- \( {n \choose m} / n^{m} \)
- \( {{m+n-1} \choose m} / m^{n} \)
- \( {{m+n-1} \choose m-1} / n^{m} \)

combination

Answer: is \( {n \choose m} / n^{m} \)

ISI MStat 2018 PSA Problem 13

A First Course in Probability by Sheldon Ross

What is the total number of functions from \({1,2, \ldots, m}\) to \({1,2, \ldots, n}\) where (n>m)

You have to choose \(m\) numbers among \({1,2, \ldots, n}\) and assign it to the \({1,2, \ldots, m}\)

For each element of \({1,2, \ldots, m}\), there are \(n\) options from \({1,2, \ldots, n}\).

Hence \(n^m \) number of functions .

\(f(i) = a_i \)

The number of ways to select Select \(m\) elements among \({1,2, \ldots, n}\) is the same as the number of strictly ascending subsequences of length m taken from 1, 2, 3, ..., n, which is the same as the number of subsets of size m taken from {1,2,3,…,n}, which is \( {n \choose m} \) .

Hence the probability that it will be strictly increasing \( \frac{ {n \choose m} }{ n^{m} } \)

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