What is the intuition behind the resolution method for this problem?
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I'm trying to understand the intuition behind the way we proceed to calculate how many different 4-digit numbers can we build using digits from $1$ to $9$, such that there is exactly one digit repeated, so for example $1982$ or $6333$ are not valid, but $2942$ and $7721$ are. To find the result, we calculate
$$9cdot text{C}^4_2 cdot text{P}^8_2 = 9binom{4}{2} cdot frac{8!}{6!} = 3024$$
Now when I look at $text{C}^4_2$ I can't think in other terms than "amount of possible subsets of size 2 that we can build from a set of 4 elements", so I really can't get the intuition of multiplying this by "the amount of different size-2 permutations we can create from a set of eight elements, without repetition". I would appreciate someone explains to me the logic of this operation, because I don't like to do mathematics mechanically and without understanding what I'm doing. Thank you.
combinatorics statistics permutations combinations
asked Nov 26 at 10:17
torito verdejo
725
add a comment |
up vote
0
down vote
favorite
I'm trying to understand the intuition behind the way we proceed to calculate how many different 4-digit numbers can we build using digits from $1$ to $9$, such that there is exactly one digit repeated, so for example $1982$ or $6333$ are not valid, but $2942$ and $7721$ are. To find the result, we calculate
$$9cdot text{C}^4_2 cdot text{P}^8_2 = 9binom{4}{2} cdot frac{8!}{6!} = 3024$$
Now when I look at $text{C}^4_2$ I can't think in other terms than "amount of possible subsets of size 2 that we can build from a set of 4 elements", so I really can't get the intuition of multiplying this by "the amount of different size-2 permutations we can create from a set of eight elements, without repetition". I would appreciate someone explains to me the logic of this operation, because I don't like to do mathematics mechanically and without understanding what I'm doing. Thank you.
combinatorics statistics permutations combinations
asked Nov 26 at 10:17
torito verdejo
725
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm trying to understand the intuition behind the way we proceed to calculate how many different 4-digit numbers can we build using digits from $1$ to $9$, such that there is exactly one digit repeated, so for example $1982$ or $6333$ are not valid, but $2942$ and $7721$ are. To find the result, we calculate
$$9cdot text{C}^4_2 cdot text{P}^8_2 = 9binom{4}{2} cdot frac{8!}{6!} = 3024$$
Now when I look at $text{C}^4_2$ I can't think in other terms than "amount of possible subsets of size 2 that we can build from a set of 4 elements", so I really can't get the intuition of multiplying this by "the amount of different size-2 permutations we can create from a set of eight elements, without repetition". I would appreciate someone explains to me the logic of this operation, because I don't like to do mathematics mechanically and without understanding what I'm doing. Thank you.
combinatorics statistics permutations combinations
asked Nov 26 at 10:17
torito verdejo
725
I'm trying to understand the intuition behind the way we proceed to calculate how many different 4-digit numbers can we build using digits from $1$ to $9$, such that there is exactly one digit repeated, so for example $1982$ or $6333$ are not valid, but $2942$ and $7721$ are. To find the result, we calculate
$$9cdot text{C}^4_2 cdot text{P}^8_2 = 9binom{4}{2} cdot frac{8!}{6!} = 3024$$
Now when I look at $text{C}^4_2$ I can't think in other terms than "amount of possible subsets of size 2 that we can build from a set of 4 elements", so I really can't get the intuition of multiplying this by "the amount of different size-2 permutations we can create from a set of eight elements, without repetition". I would appreciate someone explains to me the logic of this operation, because I don't like to do mathematics mechanically and without understanding what I'm doing. Thank you.
combinatorics statistics permutations combinations
combinatorics statistics permutations combinations
asked Nov 26 at 10:17
torito verdejo
725
asked Nov 26 at 10:17
torito verdejo
725
asked Nov 26 at 10:17
torito verdejo
725
asked Nov 26 at 10:17
torito verdejo
725
asked Nov 26 at 10:17
torito verdejo
725
725
add a comment |
add a comment |
1 Answer
1
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oldest
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2
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accepted
- Choose 1 digit out of $1, ldots, 9$: $color{blue}{9}$ possibilities
- Choose $2$ places out of $4$ to place this digit twice: $color{blue}{binom{4}{2}}$ possibilities
- Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $color{blue}{8cdot 7 = P_2^8}$ possibilities
All together
$$color{blue}{9cdot binom{4}{2} cdot 8 cdot 7}$$
answered Nov 26 at 10:21
trancelocation
8,7121520
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
- Choose 1 digit out of $1, ldots, 9$: $color{blue}{9}$ possibilities
- Choose $2$ places out of $4$ to place this digit twice: $color{blue}{binom{4}{2}}$ possibilities
- Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $color{blue}{8cdot 7 = P_2^8}$ possibilities
All together
$$color{blue}{9cdot binom{4}{2} cdot 8 cdot 7}$$
answered Nov 26 at 10:21
trancelocation
8,7121520
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
add a comment |
up vote
2
down vote
accepted
- Choose 1 digit out of $1, ldots, 9$: $color{blue}{9}$ possibilities
- Choose $2$ places out of $4$ to place this digit twice: $color{blue}{binom{4}{2}}$ possibilities
- Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $color{blue}{8cdot 7 = P_2^8}$ possibilities
All together
$$color{blue}{9cdot binom{4}{2} cdot 8 cdot 7}$$
answered Nov 26 at 10:21
trancelocation
8,7121520
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
- Choose 1 digit out of $1, ldots, 9$: $color{blue}{9}$ possibilities
- Choose $2$ places out of $4$ to place this digit twice: $color{blue}{binom{4}{2}}$ possibilities
- Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $color{blue}{8cdot 7 = P_2^8}$ possibilities
All together
$$color{blue}{9cdot binom{4}{2} cdot 8 cdot 7}$$
answered Nov 26 at 10:21
trancelocation
8,7121520
- Choose 1 digit out of $1, ldots, 9$: $color{blue}{9}$ possibilities
- Choose $2$ places out of $4$ to place this digit twice: $color{blue}{binom{4}{2}}$ possibilities
- Fill the remaining $2$ places with $2$ different digits from the remaining $8$ digits: $color{blue}{8cdot 7 = P_2^8}$ possibilities
All together
$$color{blue}{9cdot binom{4}{2} cdot 8 cdot 7}$$
answered Nov 26 at 10:21
trancelocation
8,7121520
answered Nov 26 at 10:21
trancelocation
8,7121520
answered Nov 26 at 10:21
trancelocation
8,7121520
answered Nov 26 at 10:21
trancelocation
8,7121520
8,7121520
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
add a comment |
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Oh, now I get it, and I also get why we are multiplying everything : we do so to take into account every possible position of the two other digits, for each of the possible positions where the repeated digit can be placed, right? Thank you.
– torito verdejo
Nov 26 at 10:31
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
Exactly. Basis for the multiplication here is the so called "multiplication principle" (en.wikipedia.org/wiki/Rule_of_product ) as a basic counting rule in combinatorics.
– trancelocation
Nov 26 at 10:35
add a comment |
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