Mixing
Permutations with repetition leading to fraction answer
$begingroup$
I have the following scenario where I need to find the number of unique 4 digit numbers using the following 6 digits:
4,4,4,4,6,6
According to my understanding this can be derived by the following calculation:
6P4 / 4! . 2!
However this give the following answer 7.5?
Any help will be much appreciated.
Thanks
combinatorics permutations
edited Jan 7 at 0:26
N. F. Taussig
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
$endgroup$
add a comment |
$begingroup$
I have the following scenario where I need to find the number of unique 4 digit numbers using the following 6 digits:
4,4,4,4,6,6
According to my understanding this can be derived by the following calculation:
6P4 / 4! . 2!
However this give the following answer 7.5?
Any help will be much appreciated.
Thanks
combinatorics permutations
edited Jan 7 at 0:26
N. F. Taussig
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
$endgroup$
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
2
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00
add a comment |
$begingroup$
I have the following scenario where I need to find the number of unique 4 digit numbers using the following 6 digits:
4,4,4,4,6,6
According to my understanding this can be derived by the following calculation:
6P4 / 4! . 2!
However this give the following answer 7.5?
Any help will be much appreciated.
Thanks
combinatorics permutations
edited Jan 7 at 0:26
N. F. Taussig
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
$endgroup$
I have the following scenario where I need to find the number of unique 4 digit numbers using the following 6 digits:
4,4,4,4,6,6
According to my understanding this can be derived by the following calculation:
6P4 / 4! . 2!
However this give the following answer 7.5?
Any help will be much appreciated.
Thanks
combinatorics permutations
combinatorics permutations
edited Jan 7 at 0:26
N. F. Taussig
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
edited Jan 7 at 0:26
N. F. Taussig
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
edited Jan 7 at 0:26
N. F. Taussig
44k93356
edited Jan 7 at 0:26
N. F. Taussig
44k93356
edited Jan 7 at 0:26
N. F. Taussig
44k93356
44k93356
asked Jan 6 at 23:10
Ponniah Ponniah
61
asked Jan 6 at 23:10
Ponniah Ponniah
61
asked Jan 6 at 23:10
Ponniah Ponniah
61
61
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
2
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00
add a comment |
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
2
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
2
2
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:
- AAAA which constitutes a single solution
- AAAB and permutations ${}_4 C_1=4$
- AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.
There is a problem with the model in the link you gave:
$$
frac{{}_n P_r}{a!b!cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.
This is why the figure:
$$
frac{{}_6 P_4}{4!2!}=7.5
$$
that you gave is too low.
- The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
- Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
- So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.
Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
answered Jan 6 at 23:36
StringString
13.7k32756
$endgroup$
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
|
show 3 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064504%2fpermutations-with-repetition-leading-to-fraction-answer%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:
- AAAA which constitutes a single solution
- AAAB and permutations ${}_4 C_1=4$
- AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.
There is a problem with the model in the link you gave:
$$
frac{{}_n P_r}{a!b!cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.
This is why the figure:
$$
frac{{}_6 P_4}{4!2!}=7.5
$$
that you gave is too low.
- The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
- Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
- So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.
Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
answered Jan 6 at 23:36
StringString
13.7k32756
$endgroup$
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
|
show 3 more comments
$begingroup$
This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:
- AAAA which constitutes a single solution
- AAAB and permutations ${}_4 C_1=4$
- AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.
There is a problem with the model in the link you gave:
$$
frac{{}_n P_r}{a!b!cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.
This is why the figure:
$$
frac{{}_6 P_4}{4!2!}=7.5
$$
that you gave is too low.
- The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
- Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
- So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.
Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
answered Jan 6 at 23:36
StringString
13.7k32756
$endgroup$
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
|
show 3 more comments
$begingroup$
This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:
- AAAA which constitutes a single solution
- AAAB and permutations ${}_4 C_1=4$
- AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.
There is a problem with the model in the link you gave:
$$
frac{{}_n P_r}{a!b!cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.
This is why the figure:
$$
frac{{}_6 P_4}{4!2!}=7.5
$$
that you gave is too low.
- The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
- Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
- So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.
Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
answered Jan 6 at 23:36
StringString
13.7k32756
$endgroup$
This corresponds to counting the number of 4-letter words formed from AAAABB. We have three cases:
- AAAA which constitutes a single solution
- AAAB and permutations ${}_4 C_1=4$
- AABB and permutations ${}_4 C_2=6$
This gives us a total of $11$ words.
There is a problem with the model in the link you gave:
$$
frac{{}_n P_r}{a!b!cdots}
$$
where $a,b,...$ are the number of repetitions of the different items.
The problem is that this model breaks down whenever the number of unchosen items $n-r$ is greater than $2$ while there is at least one repeated item in the list. The reason for this is that the figures in the denominator also count permutations of the $n-r$ items that where NOT chosen, but the numerator only counts permutations of the $r$ chosen items.
This is why the figure:
$$
frac{{}_6 P_4}{4!2!}=7.5
$$
that you gave is too low.
- The numerator ${}_6 P_4$ counts $4444$ as $4!$ permutations.
- Whereas the denominator counts $4444$ not choosing $66$ as $4!2!$ permutations.
- So instead of contributing one arrangement, the case $4444$ contributes only $1/2$ arrangement.
Similarly, the numerator counts $4466$ as ${}_4 P_2 2!=4!$ permutations, but divides them by $4!2!$ counting only half an arrangement. The same can be said for each of the ${}_4 P_4/(2!2!)=6$ rearrangements of $4466$. So those $6$ cases are only counted as half as many, namely $3$.
Conclusion: The method given in the link is useless when the difference between $n$ and $r$ is greater than $1$.
answered Jan 6 at 23:36
StringString
13.7k32756
edited Jan 8 at 10:07
answered Jan 6 at 23:36
StringString
13.7k32756
answered Jan 6 at 23:36
StringString
13.7k32756
answered Jan 6 at 23:36
StringString
13.7k32756
13.7k32756
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
|
show 3 more comments
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
$begingroup$
Thanks everyone for the replies. So it looks like the generic formula found in books for permutation with repetition cannot be applied for all the cases.
$endgroup$
– Ponniah
Jan 7 at 8:35
1
1
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
Please check the below link where it’s been advised to divide by number of factorial of the repeated item. ... ck12.org/probability/permutations-with-repetition/lesson/…
$endgroup$
– Ponniah
Jan 8 at 6:44
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: Ah, I see the problem! Perhaps you could include that link in the OP, and I will try my best to address is in my answer.
$endgroup$
– String
Jan 8 at 8:32
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
@Ponniah: I just added a large section on the problem with the method given in the link. Feel free to ask for clarifications!
$endgroup$
– String
Jan 8 at 9:02
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
$begingroup$
Thanks a lot for this, it’s very clear now!!!
$endgroup$
– Ponniah
Jan 8 at 10:06
|
show 3 more comments
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3064504%2fpermutations-with-repetition-leading-to-fraction-answer%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
@String Thanks for pointing out that I'd misread the question!
$endgroup$
– timtfj
Jan 6 at 23:46
$begingroup$
@timtfj: You're welcome!
$endgroup$
– String
Jan 6 at 23:46
$begingroup$
@String I should clearly post only when properly awake. Which isn't very often.
$endgroup$
– timtfj
Jan 6 at 23:48
2
$begingroup$
Dividing by $4!$ assume you have all four $4$s in you number and dividing by $2!$ assumes you have all two $6$s. As you only have $4$ numbers, this just isn't the correct model to use.
$endgroup$
– fleablood
Jan 7 at 1:00