Post History
#6: Post edited
by
DNB
·
2021-07-09T04:33:05Z (almost 3 years ago)
Please see the embolded phrase below. How can you explain to a 9 year old why you1. must divide by $c$?2. can't subtract by $c$?>### 1.4.2 Adjusting for overcounting>In many counting problems, it is not easy to directly count each possibility onceand only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactlydouble-counted each possibility, we can divide by 2 to get the correct count. We callthis _adjusting for overcounting_.Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
- Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
- 1. How can I deduce which operation ought fill in the red blank beneath?
- 2. Why can't it be subtraction?
- ### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
- >**Quandary:** How many ways can I give gifts to people, if nobody receives gift?
- >
- >**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
- >
- >For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for people, it’s just .
- >
- >So, for giving 3 gifts, there are variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}\text{_____}$ by all the redundancies. [...]
#5: Post edited
by
DNB
·
2021-07-09T04:31:29Z (almost 3 years ago)
Please see the embolded phrase below. How can you explain to a 10 year old why1. you must divide by $c$?2. you can't subtract by $c$?- >### 1.4.2 Adjusting for overcounting
- >In many counting problems, it is not easy to directly count each possibility once
- and only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactly
- double-counted each possibility, we can divide by 2 to get the correct count. We call
- this _adjusting for overcounting_.
- Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
- Please see the embolded phrase below. How can you explain to a 9 year old why you
- 1. must divide by $c$?
- 2. can't subtract by $c$?
- >### 1.4.2 Adjusting for overcounting
- >In many counting problems, it is not easy to directly count each possibility once
- and only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactly
- double-counted each possibility, we can divide by 2 to get the correct count. We call
- this _adjusting for overcounting_.
- Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
#4: Post edited
by
DNB
·
2021-07-09T04:24:47Z (almost 3 years ago)
- Please see the embolded phrase below. How can you explain to a 10 year old why
- 1. you must divide by $c$?
- 2. you can't subtract by $c$?
>###1.4.2 Adjusting for overcounting- >In many counting problems, it is not easy to directly count each possibility once
- and only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactly
- double-counted each possibility, we can divide by 2 to get the correct count. We call
- this _adjusting for overcounting_.
- Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
- Please see the embolded phrase below. How can you explain to a 10 year old why
- 1. you must divide by $c$?
- 2. you can't subtract by $c$?
- >### 1.4.2 Adjusting for overcounting
- >In many counting problems, it is not easy to directly count each possibility once
- and only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactly
- double-counted each possibility, we can divide by 2 to get the correct count. We call
- this _adjusting for overcounting_.
- Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
#3: Post edited
by
DNB
·
2021-07-09T04:23:55Z (almost 3 years ago)
Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.1. How can I deduce which operation ought fill in the red blank beneath?2. Why can't it be subtraction?### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):>**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?>>**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.>>For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.>>So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}\text{_____}$ by all the redundancies. [...]
- Please see the embolded phrase below. How can you explain to a 10 year old why
- 1. you must divide by $c$?
- 2. you can't subtract by $c$?
- >###1.4.2 Adjusting for overcounting
- >In many counting problems, it is not easy to directly count each possibility once
- and only once. If, however, we are able to count each possibility exactly c times for some c, then we can adjust **by dividing by $c$**. For example, if we have exactly
- double-counted each possibility, we can divide by 2 to get the correct count. We call
- this _adjusting for overcounting_.
- Blitzstein. *Introduction to Probability* (2019 2 ed). pp. 14-15.
#2: Post edited
by
r~~
·
2021-01-02T20:42:30Z (over 3 years ago)
Fix TeX formatting
How can I deduce which operation removes redundacies?
- Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
- 1. How can I deduce which operation ought fill in the red blank beneath?
- 2. Why can't it be subtraction?
- ### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
- >**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?
- >
- >**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
- >
- >For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.
- >
>So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}{\text{_____ by all the redundancies}}$. [...]
- Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
- 1. How can I deduce which operation ought fill in the red blank beneath?
- 2. Why can't it be subtraction?
- ### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
- >**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?
- >
- >**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
- >
- >For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.
- >
- >So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}\text{_____}$ by all the redundancies. [...]
#1: Initial revision
by
DNB
·
2020-12-30T03:11:56Z (over 3 years ago)
How can I deduce which operation removes redundacies?
Please don't answer by working backwards from the answer, or by appealing to arithmetic. Act as if you're learning this for the first time.
1. How can I deduce which operation ought fill in the red blank beneath?
2. Why can't it be subtraction?
### [I shortened the original explanation](https://betterexplained.com/articles/easy-permutations-and-combinations/):
>**Quandary:** How many ways can I give $3$ gifts to $8$ people, if nobody receives $>1$ gift?
>
>**Explanation:** For combinations, order doesn’t matter. This raises an interesting detail of redundancies.
>
>For a moment, let's figure out how many ways we can rearrange 3 people. We've 3 choices for the first person, 2 for the second, and only 1 for the last. So we have $3 · 2 · 1$ ways to re-arrange 3 people. But this is a permutation! If you want to know the number of arrangements for $N$ people, it’s just $N!$.
>
>So, for giving 3 gifts, there are $3! \; (= 6)$ variations for every choice we pick. To calculate how many combinations, just create all the permutations and $\color{red}{\text{_____ by all the redundancies}}$. [...]