Post History
#6: Post edited
by
DNB
·
2021-07-28T04:05:35Z (over 3 years ago)
- 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](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_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.
- 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](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_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.
#5: Post edited
by
DNB
·
2021-07-09T07:53:24Z (over 3 years ago)
Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots {(n-k+1)}\color{red}{(n - k)}$?
- Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots [(n-(k - 1)]\color{red}{(n - k)}$?
I know that $\color{red}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as $n(n-1) \dots \color{red}{(n-k+1)}(n - k)$. I bungle by adding the unnecessary and wrong $n - k$, probably because I affiliate choosing K objects with $n - k$!How can I rectify my [Fence Post Error](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error)? How can I intuitively remember to stop at \color{red}{(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 $n(n-1) \dots \color{red}{(n-k+1)}$- 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.
- 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](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_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.
#4: Post edited
by
DNB
·
2021-07-09T07:49:44Z (over 3 years ago)
Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots {\color{red}{(n-k+1)}}(n - k)$?
- Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots {(n-k+1)}\color{red}{(n - k)}$?
#3: Post edited
by
DNB
·
2021-07-09T04:10:05Z (over 3 years ago)
Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots \color{red}{(n-k+1)}(n - (k - 1))$?
- Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots {\color{red}{(n-k+1)}}(n - k)$?
#2: Post edited
by
DNB
·
2021-07-09T04:09:19Z (over 3 years ago)
- I know that $\color{red}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as $n(n-1) \dots \color{red}{(n-k+1)}(n - k)$. I bungle by adding the unnecessary and wrong $n - k$, probably because I affiliate choosing K objects with $n - k$!
- How can I rectify my [Fence Post Error](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error)? How can I intuitively remember to stop at \color{red}{(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 $n(n-1) \dots \color{red}{(n-k+1)}$
- 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.
- I know that $\color{red}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as $n(n-1) \dots \color{red}{(n-k+1)}(n - k)$. I bungle by adding the unnecessary and wrong $n - k$, probably because I affiliate choosing K objects with $n - k$!
- How can I rectify my [Fence Post Error](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error)? How can I intuitively remember to stop at \color{red}{(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 $n(n-1) \dots \color{red}{(n-k+1)}$
- 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
·
2021-07-09T04:08:51Z (over 3 years ago)
Intuitively, if you pick k out of n objects singly without replacement, why's the number of possible outcomes NOT $n(n-1) \dots \color{red}{(n-k+1)}(n - (k - 1))$?
I know that $\color{red}{(n-k+1)} \equiv (n - (k - 1))$. But whenever I contemplate choosing k from n objects singly without replacement, I keep muffing the number of possible outcomes as $n(n-1) \dots \color{red}{(n-k+1)}(n - k)$. I bungle by adding the unnecessary and wrong $n - k$, probably because I affiliate choosing K objects with $n - k$!
How can I rectify my [Fence Post Error](https://en.wikipedia.org/wiki/Off-by-one_error#Fencepost_error)? How can I intuitively remember to stop at \color{red}{(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 $n(n-1) \dots \color{red}{(n-k+1)}$
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.