Q&A

# 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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