How many permutations of 7 digits, 4 digits the same, & 3 different from the 4 but also the same
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In other words, how many permutations are available for 4A3B, e.g., 3333999, 3939339.
I guess this could be restated as how many permutations are available from X digits, with Y the same, and Z the same but where Y<>Z. But really I can do the generalization. I'm having a hard time visualizing this. I'm more interesting in thought process than the actual answer.
I'm assuming permutations like 0000111 are included in the count.
probability
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
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add a comment |
$begingroup$
In other words, how many permutations are available for 4A3B, e.g., 3333999, 3939339.
I guess this could be restated as how many permutations are available from X digits, with Y the same, and Z the same but where Y<>Z. But really I can do the generalization. I'm having a hard time visualizing this. I'm more interesting in thought process than the actual answer.
I'm assuming permutations like 0000111 are included in the count.
probability
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
$endgroup$
$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
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let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46
add a comment |
$begingroup$
In other words, how many permutations are available for 4A3B, e.g., 3333999, 3939339.
I guess this could be restated as how many permutations are available from X digits, with Y the same, and Z the same but where Y<>Z. But really I can do the generalization. I'm having a hard time visualizing this. I'm more interesting in thought process than the actual answer.
I'm assuming permutations like 0000111 are included in the count.
probability
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
$endgroup$
In other words, how many permutations are available for 4A3B, e.g., 3333999, 3939339.
I guess this could be restated as how many permutations are available from X digits, with Y the same, and Z the same but where Y<>Z. But really I can do the generalization. I'm having a hard time visualizing this. I'm more interesting in thought process than the actual answer.
I'm assuming permutations like 0000111 are included in the count.
probability
probability
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
asked Dec 27 '18 at 22:33
GLearnerGLearner
141
141
$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
$begingroup$
let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46
add a comment |
$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
$begingroup$
let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46
$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
$begingroup$
let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46
$begingroup$
let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Basic idea: There are $(Y+Z)!$ permutations of the digits. However $Y!$ and $Z!$ are the same arrangements for each digit respectively, so the net result is $frac{(Y+Z)!}{Y!Z!}$
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
$endgroup$
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
add a comment |
$begingroup$
There are ${10choose 1}$ ways to choose the first category of digits, then there are ${9choose 1}$ ways to choose the second category of digits. Finally, there are $frac{7!}{4!cdot 3!}$ ways to permute the digits:
$${10choose 1}{9choose 1}frac{7!}{4!cdot 3!}.$$
In general, for the $X$-digit number with $Y$ the same and $Z$ the same, but $Yne Z$:
$${10choose 1}{9choose 1}cdot frac{X!}{Y!Z!}.$$
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Basic idea: There are $(Y+Z)!$ permutations of the digits. However $Y!$ and $Z!$ are the same arrangements for each digit respectively, so the net result is $frac{(Y+Z)!}{Y!Z!}$
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
$endgroup$
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
add a comment |
$begingroup$
Basic idea: There are $(Y+Z)!$ permutations of the digits. However $Y!$ and $Z!$ are the same arrangements for each digit respectively, so the net result is $frac{(Y+Z)!}{Y!Z!}$
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
$endgroup$
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
add a comment |
$begingroup$
Basic idea: There are $(Y+Z)!$ permutations of the digits. However $Y!$ and $Z!$ are the same arrangements for each digit respectively, so the net result is $frac{(Y+Z)!}{Y!Z!}$
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
$endgroup$
Basic idea: There are $(Y+Z)!$ permutations of the digits. However $Y!$ and $Z!$ are the same arrangements for each digit respectively, so the net result is $frac{(Y+Z)!}{Y!Z!}$
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
answered Dec 27 '18 at 22:45
herb steinbergherb steinberg
3,1832311
3,1832311
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
add a comment |
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
I may have been unclear, but that’s not what I’m asking. I believe your answer gives me how many possibilities for a given selection within the 3A4B or XAYB grouping. What I’m asking for is all possibilities across all selections within the 3A3B or XAYB groupings. I tried 10x2x2x2x2x2x2 but that gets me close but I’m missing something. I think I’m double counting.
$endgroup$
– GLearner
Dec 27 '18 at 23:47
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
$begingroup$
The answer I gave you is the total number of different possibilities. Are you asking for the number of arrangements where all look the same? That would be $Y!Z!$. You need to clarify what you are looking for.
$endgroup$
– herb steinberg
Dec 28 '18 at 1:01
add a comment |
$begingroup$
There are ${10choose 1}$ ways to choose the first category of digits, then there are ${9choose 1}$ ways to choose the second category of digits. Finally, there are $frac{7!}{4!cdot 3!}$ ways to permute the digits:
$${10choose 1}{9choose 1}frac{7!}{4!cdot 3!}.$$
In general, for the $X$-digit number with $Y$ the same and $Z$ the same, but $Yne Z$:
$${10choose 1}{9choose 1}cdot frac{X!}{Y!Z!}.$$
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
$endgroup$
add a comment |
$begingroup$
There are ${10choose 1}$ ways to choose the first category of digits, then there are ${9choose 1}$ ways to choose the second category of digits. Finally, there are $frac{7!}{4!cdot 3!}$ ways to permute the digits:
$${10choose 1}{9choose 1}frac{7!}{4!cdot 3!}.$$
In general, for the $X$-digit number with $Y$ the same and $Z$ the same, but $Yne Z$:
$${10choose 1}{9choose 1}cdot frac{X!}{Y!Z!}.$$
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
$endgroup$
add a comment |
$begingroup$
There are ${10choose 1}$ ways to choose the first category of digits, then there are ${9choose 1}$ ways to choose the second category of digits. Finally, there are $frac{7!}{4!cdot 3!}$ ways to permute the digits:
$${10choose 1}{9choose 1}frac{7!}{4!cdot 3!}.$$
In general, for the $X$-digit number with $Y$ the same and $Z$ the same, but $Yne Z$:
$${10choose 1}{9choose 1}cdot frac{X!}{Y!Z!}.$$
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
$endgroup$
There are ${10choose 1}$ ways to choose the first category of digits, then there are ${9choose 1}$ ways to choose the second category of digits. Finally, there are $frac{7!}{4!cdot 3!}$ ways to permute the digits:
$${10choose 1}{9choose 1}frac{7!}{4!cdot 3!}.$$
In general, for the $X$-digit number with $Y$ the same and $Z$ the same, but $Yne Z$:
$${10choose 1}{9choose 1}cdot frac{X!}{Y!Z!}.$$
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
answered Dec 28 '18 at 10:08
farruhotafarruhota
22.2k2942
22.2k2942
add a comment |
add a comment |
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$begingroup$
Could you consider clarifying the question?
$endgroup$
– pbn990
Dec 28 '18 at 0:43
$begingroup$
let’s pretend we have a combination 3334444. This has 3 digits that are the same and 4 that are the same. Total 7 digits. You can produce 35 permutations of this, for example 4343344 is one. What are the total possibilities of all such 3A4B (e.g., 0001111) permutations for all 0-9 digits. (4+3)!/(4!3!) = 35 but that only covers one possiblity.
$endgroup$
– GLearner
Dec 28 '18 at 3:46