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Given 10 digits, how many ways can they be arranged so that two odds cannot be adjacent?
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Given $10$ digits, where each digit can be an integer from $0$ to $9$, how can I determine the number of ways to arrange the numbers so that two odds are not adjacent?
Repetition of digits is not allowed.
So far, I have figured out the total number of possibilities: $$10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1 = 10!$$
Then I had planned to subtract the number of bad possibilities from the total number of possibilities.$$10! - X$$ Where $X$ is all the bad possibilities, which means $X$ is all the possibilities where two odds could be next to each other in the $10$ digits.
I know that for each number, $5$ odds can be selected, how can I use this information to figure out the answer to the question?
combinatorics
asked 1 hour ago
Matt Hough
945
add a comment |Â
up vote
1
down vote
favorite
Given $10$ digits, where each digit can be an integer from $0$ to $9$, how can I determine the number of ways to arrange the numbers so that two odds are not adjacent?
Repetition of digits is not allowed.
So far, I have figured out the total number of possibilities: $$10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1 = 10!$$
Then I had planned to subtract the number of bad possibilities from the total number of possibilities.$$10! - X$$ Where $X$ is all the bad possibilities, which means $X$ is all the possibilities where two odds could be next to each other in the $10$ digits.
I know that for each number, $5$ odds can be selected, how can I use this information to figure out the answer to the question?
combinatorics
asked 1 hour ago
Matt Hough
945
If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given $10$ digits, where each digit can be an integer from $0$ to $9$, how can I determine the number of ways to arrange the numbers so that two odds are not adjacent?
Repetition of digits is not allowed.
So far, I have figured out the total number of possibilities: $$10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1 = 10!$$
Then I had planned to subtract the number of bad possibilities from the total number of possibilities.$$10! - X$$ Where $X$ is all the bad possibilities, which means $X$ is all the possibilities where two odds could be next to each other in the $10$ digits.
I know that for each number, $5$ odds can be selected, how can I use this information to figure out the answer to the question?
combinatorics
asked 1 hour ago
Matt Hough
945
Given $10$ digits, where each digit can be an integer from $0$ to $9$, how can I determine the number of ways to arrange the numbers so that two odds are not adjacent?
Repetition of digits is not allowed.
So far, I have figured out the total number of possibilities: $$10 cdot 9 cdot 8 cdot 7 cdot 6 cdot 5 cdot 4 cdot 3 cdot 2 cdot 1 = 10!$$
Then I had planned to subtract the number of bad possibilities from the total number of possibilities.$$10! - X$$ Where $X$ is all the bad possibilities, which means $X$ is all the possibilities where two odds could be next to each other in the $10$ digits.
I know that for each number, $5$ odds can be selected, how can I use this information to figure out the answer to the question?
combinatorics
combinatorics
asked 1 hour ago
Matt Hough
945
asked 1 hour ago
Matt Hough
945
asked 1 hour ago
Matt Hough
945
asked 1 hour ago
Matt Hough
945
asked 1 hour ago
Matt Hough
945
945
If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago
add a comment |Â
If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago
If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
There are six different admissible assignments of odd (O) and even (e) numbers:
OeOeOeOeOe
OeOeOeOeeO
OeOeOeeOeO
OeOeeOeOeO
OeeOeOeOeO
eOeOeOeOeO
For these arrangements of parity there are $5!$ ways of fixing the odd numbers, and the same number of ways to fix the even numbers. Thus there are $6×5!×5!=86400$ ways.
answered 53 mins ago
Parcly Taxel
36.6k137095
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
add a comment |Â
up vote
1
down vote
Let $O$ and $E$ be the odd and even numbers respectively.
The odd digits are $1,3,5,7,9$.
The even digits are $0,2,4,6,8$.
$_E_E_E_E_E_$
If you are filling the odd numbers in any $5$ blank spaces yiels a required number.
There are $6$ blank spaces. So the number of ways to select $5$ spaces among $6$ spaces $=6C5 =6$.
Number of possible shuffling on $5$ odd number is $=5!=120$.
Again,Number of possible shuffling on $5$ even number is $=5!=120$.
Thus total number of possibility is $=6×120×120=86400$.
answered 43 mins ago
Avinash N
487411
Is there a reason the image has 11 spots? (_E_E_E_E_E_)
– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
There are six different admissible assignments of odd (O) and even (e) numbers:
OeOeOeOeOe
OeOeOeOeeO
OeOeOeeOeO
OeOeeOeOeO
OeeOeOeOeO
eOeOeOeOeO
For these arrangements of parity there are $5!$ ways of fixing the odd numbers, and the same number of ways to fix the even numbers. Thus there are $6×5!×5!=86400$ ways.
answered 53 mins ago
Parcly Taxel
36.6k137095
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
add a comment |Â
up vote
2
down vote
accepted
There are six different admissible assignments of odd (O) and even (e) numbers:
OeOeOeOeOe
OeOeOeOeeO
OeOeOeeOeO
OeOeeOeOeO
OeeOeOeOeO
eOeOeOeOeO
For these arrangements of parity there are $5!$ ways of fixing the odd numbers, and the same number of ways to fix the even numbers. Thus there are $6×5!×5!=86400$ ways.
answered 53 mins ago
Parcly Taxel
36.6k137095
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
There are six different admissible assignments of odd (O) and even (e) numbers:
OeOeOeOeOe
OeOeOeOeeO
OeOeOeeOeO
OeOeeOeOeO
OeeOeOeOeO
eOeOeOeOeO
For these arrangements of parity there are $5!$ ways of fixing the odd numbers, and the same number of ways to fix the even numbers. Thus there are $6×5!×5!=86400$ ways.
answered 53 mins ago
Parcly Taxel
36.6k137095
There are six different admissible assignments of odd (O) and even (e) numbers:
OeOeOeOeOe
OeOeOeOeeO
OeOeOeeOeO
OeOeeOeOeO
OeeOeOeOeO
eOeOeOeOeO
For these arrangements of parity there are $5!$ ways of fixing the odd numbers, and the same number of ways to fix the even numbers. Thus there are $6×5!×5!=86400$ ways.
answered 53 mins ago
Parcly Taxel
36.6k137095
answered 53 mins ago
Parcly Taxel
36.6k137095
answered 53 mins ago
Parcly Taxel
36.6k137095
answered 53 mins ago
Parcly Taxel
36.6k137095
36.6k137095
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
add a comment |Â
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
the final one might have an issue with starting $0$ digit. Not sure if it's allowed. One would need to substract some options.
– Henno Brandsma
44 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
@HennoBrandsma it was not specified in the question that the result had to be a 10-digit number.
– Parcly Taxel
42 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
It's just 10 digits, not a 10 digit number. It's fine.
– Matt Hough
41 mins ago
add a comment |Â
up vote
1
down vote
Let $O$ and $E$ be the odd and even numbers respectively.
The odd digits are $1,3,5,7,9$.
The even digits are $0,2,4,6,8$.
$_E_E_E_E_E_$
If you are filling the odd numbers in any $5$ blank spaces yiels a required number.
There are $6$ blank spaces. So the number of ways to select $5$ spaces among $6$ spaces $=6C5 =6$.
Number of possible shuffling on $5$ odd number is $=5!=120$.
Again,Number of possible shuffling on $5$ even number is $=5!=120$.
Thus total number of possibility is $=6×120×120=86400$.
answered 43 mins ago
Avinash N
487411
Is there a reason the image has 11 spots? (_E_E_E_E_E_)
– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
add a comment |Â
up vote
1
down vote
Let $O$ and $E$ be the odd and even numbers respectively.
The odd digits are $1,3,5,7,9$.
The even digits are $0,2,4,6,8$.
$_E_E_E_E_E_$
If you are filling the odd numbers in any $5$ blank spaces yiels a required number.
There are $6$ blank spaces. So the number of ways to select $5$ spaces among $6$ spaces $=6C5 =6$.
Number of possible shuffling on $5$ odd number is $=5!=120$.
Again,Number of possible shuffling on $5$ even number is $=5!=120$.
Thus total number of possibility is $=6×120×120=86400$.
answered 43 mins ago
Avinash N
487411
Is there a reason the image has 11 spots? (_E_E_E_E_E_)
– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $O$ and $E$ be the odd and even numbers respectively.
The odd digits are $1,3,5,7,9$.
The even digits are $0,2,4,6,8$.
$_E_E_E_E_E_$
If you are filling the odd numbers in any $5$ blank spaces yiels a required number.
There are $6$ blank spaces. So the number of ways to select $5$ spaces among $6$ spaces $=6C5 =6$.
Number of possible shuffling on $5$ odd number is $=5!=120$.
Again,Number of possible shuffling on $5$ even number is $=5!=120$.
Thus total number of possibility is $=6×120×120=86400$.
answered 43 mins ago
Avinash N
487411
Let $O$ and $E$ be the odd and even numbers respectively.
The odd digits are $1,3,5,7,9$.
The even digits are $0,2,4,6,8$.
$_E_E_E_E_E_$
If you are filling the odd numbers in any $5$ blank spaces yiels a required number.
There are $6$ blank spaces. So the number of ways to select $5$ spaces among $6$ spaces $=6C5 =6$.
Number of possible shuffling on $5$ odd number is $=5!=120$.
Again,Number of possible shuffling on $5$ even number is $=5!=120$.
Thus total number of possibility is $=6×120×120=86400$.
answered 43 mins ago
Avinash N
487411
answered 43 mins ago
Avinash N
487411
answered 43 mins ago
Avinash N
487411
answered 43 mins ago
Avinash N
487411
487411
Is there a reason the image has 11 spots? (_E_E_E_E_E_)
– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
add a comment |Â
Is there a reason the image has 11 spots? (_E_E_E_E_E_)
– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
Is there a reason the image has 11 spots? (
_E_E_E_E_E_)– Matt Hough
35 mins ago
Is there a reason the image has 11 spots? (
_E_E_E_E_E_)– Matt Hough
35 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
I think, there is no reason behind it. I just think logically and post the answer.
– Avinash N
31 mins ago
add a comment |Â
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If you only care about odd or even for a second, how many ways can you distribute the five odds and the five evens so that no odds are adjacent?
– Arthur
1 hour ago
Hint: put the odds in a row (how many possibilities?) and then separate consecutives by placing an even (how many possibilities?). After that only one even number is left and must be placed. Watch out for multiple counting.
– drhab
1 hour ago