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

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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.

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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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Thanks. But what's wrong with the 3rd line of my post? My algebra's correct. (6 comments)

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

Thanks. But what's wrong with the 3rd line of my post? My algebra's correct.

Skipping 1 deleted comment.

r~~‭ wrote about 1 year ago:

I don't even know what the middle expression in that line is supposed to mean. You start at $n$, and then descend to $n - 1$, and by the end somehow... wrap around to $n + 2$, $n + 1$? What's happening in between, where the ellipsis is? How did you arrive at that as what $LHS|_{k=1}$ should represent?

DNB‭ wrote about 1 year ago:

You wrote "You started on the right track with $n(n - 1)\ldots(n - [k - 3])(n - [k - 2])(n - [k - 1])$". I just evaluated this expression at $k = 1$. Now do you understand "what the middle expression in that line is supposed to mean"?

r~~‭ wrote about 1 year ago:

$\ldots$ doesn't work like that; you can't simply substitute a variable into that kind of informal, descriptive expression without thinking about what the notation as a whole represents. It's not an algebraic expression, so the rules of algebra don't apply. $\ldots$ is a signal to look at the pattern being established to the left of the dots, and extend it until it meets what is right of the dots. But if the patterns can't meet, the expression doesn't mean anything.

DNB‭ wrote about 1 year ago:

Thanks again. 1. Can you please expound why "you can't simply substitute a variable into that kind of informal, descriptive expression"? 2. "It's not an algebraic expression, so the rules of algebra don't apply" How isn't this an algebraic expression? This is a product of variables!

r~~‭ wrote about 1 year ago:

An algebraic expression is made exclusively of numbers, variables, and algebraic operations. $\ldots$ is none of these. This really isn't any more complicated than that. Doing algebra with informal notations can get you into trouble if you don't keep in mind what the informal notations mean. This is an example of that kind of trouble; your substitution left you with an expression that doesn't mean anything.

Skipping 1 deleted comment.

r~~‭ wrote about 1 year ago:

Nope. As written, this is a complete answer to your original question. You asked a new question here in the comments.

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