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Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots [(n-(k - 1)]\color{red}{(n - k)}$?

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I know that $\color{limegreen}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as $n(n-1) \dots \color{limegreen}(n-k+1)\color{red}{(n - k)}$. I bungled by adding the unnecessary and wrong $\color{red}{(n - k)}$, probably because I affiliated choosing K objects with $\color{red}{(n - k)}$!

How can I rectify my Fence Post Error? How can I intuitively remember to stop at $\color{limegreen}{(n-(k-1)}$?

Theorem 1.4.8 (Sampling without replacement).

Consider n objects and making k choices from them, one at a time without replacement (i.e., choosing a certain object precludes it from being chosen again). Then there are $n(n-1) \dots \color{limegreen}{(n-k+1)}$ possible outcomes for $1 \le k \le n$, and 0 possibilities for $k > n$ (where order matters). By convention, $n(n-1) \dots {\color{limegreen}{(n-k+1)}} = n$ for k = 1.

This result also follows directly from the multiplication rule: each sampled ball is again a sub-experiment, and the number of possible outcomes decreases by 1 each time. Note that for sampling k out of n objects without replacement, we need $k \le n$, whereas in sampling with replacement the objects are inexhaustible.

Blitzstein. Introduction to Probability (2019 2 ed). p. 12.

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Each term of that product gives the size of the pool when you pick an object out – it's the number of possible items you can pick out at that moment, if you're picking them one at a time.

Once you've taken $k$ items out of a pool of $n$ items, there are $n - k$ objects left over. This means that, when you were picking the last item out, your last item was also in the pool, giving a final term of $(n - k) + 1$.


I find the easiest way to get an intuition for something like this is to derive it from first principles. Mathematics only works for real-world problems when there's a relation between the problem and the mathematics, but it goes the other way, too: if some mathematics solves a problem, and you try to solve the problem from scratch, you'll end up doing something equivalent to that mathematics. If your solution is intuitive to you, then that intuition also applies to the mathematics.

Ultimately, the solution to a “this is not intuitive” problem is specific to the way you think. That also means such questions probably aren't appropriate for this site; consider asking a teacher instead.

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