# m-element random sample being equally likely …(CLRS 5.3-7)? [closed]

I am trying to understand the following solution to CLRS 5.3-7: http://clrs.skanev.com/05/03/07.html Question description is on the page. I understood the part where m-element subset is constructed out of m-1 element subset when the m-element subset includes the last element n. But why does the probability of randomly choosing m-1 elements from a set of n-1 elements differs in the case when we don't include the last element? Doesn't the invariant state that we choose m-1 elements at random from n-1 elements?

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Here is the recursive algorithm for selecting a random set $S \in \binom{[n]}{m}$ (the set of all $m$-subsets of $[n] = \{1,\ldots,n\}$):

1. If $m = 0$, return the empty set.
2. Generate a random set $S \in \binom{[n-1]}{m-1}$ recursively.
3. Let $i \in [n]$ be chosen uniformly at random.
4. If $i \notin S$, return $S \cup \{i\}$.
5. If $i \in S$, return $S \cup \{n\}$.

To show that this works, we prove that the probability to get any set in $\binom{[n]}{m}$ is exactly $1/\binom{n}{m}$, by induction on $n$. This is clear when $m = 0$, so assume $m > 0$. Let $S \in \binom{n}{m}$ be given. We consider two cases: $n \notin S$ and $n \in S$.

Case 1, $n \notin S$. For each $i \in S$, the probability that the recursive step generates $S \setminus \{i\}$ is $1/\binom{n-1}{m-1}$, and so in that case $S$ is produced with probability $1/n$. In total, the probability is $$\frac{m}{n} \frac{1}{\binom{n-1}{m-1}} = \frac{1}{\binom{n}{m}}.$$

Case 2, $n \in S$. In that case, the recursive step must generate $S \setminus \{n\}$, and then $S$ is produced with probability $m/n$ (since we either hit $n$ or any element in $S \setminus \{n\}$), obtaining the same expression as above.