Post History
#2: Post edited
by
DNB
·
2021-07-09T07:47:58Z (over 3 years ago)
Please see the boldened sentence below. I write out the LHS $= n(n-1) \dots (n-[k-3])(n-[k-2])(n-[k-1])$. 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.
- 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.
#1: Initial revision
by
DNB
·
2021-07-09T04:21:14Z (over 3 years ago)
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 $= n(n-1) \dots (n-[k-3])(n-[k-2])(n-[k-1])$. 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.