## Parent

# If k = 1, why $n(n-1) \dots \color{red}{(n-k+1)} = n$?

Please see the boldened sentence below. I write out the LHS $= n(n-1) \dots (n-[k-3])(n-[k-2])\color{red}{(n-[k-1])}$. Then $LHS|_ {k = 1} = n(n-1) \dots (n+2)(n+1) \neq n$.

## 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 $n(n-1) \dots \color{red}{(n-k+1)}$ possible outcomes for $1 \le k \le n$, and 0 possibilities for $k > n$ (where order matters).By convention, $n(n-1) \dots {\color{red}{(n-k+1)}} = n$ 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.

## Post

You started on the right track with $n(n - 1)\ldots(n - [k - 3])(n - [k - 2])(n - [k - 1])$.

What you should have noticed is that there are $k$ terms in that product. The first term subtracts 0 from $n$, the second term subtracts 1, and so on, with the $i$th term subtracting $i - 1$, all the way up to the $k$th term subtracting $k - 1$.

So the logical extension of this pattern to $k = 1$ is a product with only one term, the initial one, which subtracts 0 from $n$... which is just $n$.

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