Number of ways of forming $8$-digit odd number using only the digits $0, 0, 2, 2, 3, 3, 4, 5$ The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Smallest value of n to form 900 n-digit numbers using given digitsFind the sum of all 4 digit numbers which are formed by the digits 1,2,5,6?How many $3$-digit numbers can be formed so that the sum of two digits will be equal to the third digit?Eight digit number is formed using all the digits: $1,1,2,2,3,3,4,5$How many numbers of $7$ digits can be formed with the digit $0,1,1,5,6,6,6$.How many odd numbers of $5$ digits can be formed with the digits $0,2,3,4,5$ without repetition of any digit?Sum of all 5 digit numbers using the digits 1,2,3,4,5 at most once.How many $5$-digit positive even numbers can be formed by using all of the digits $1$, $2$, $3$ and $6$?How many 4 digit numbers can be formed from digits 0 to 9 without repetition which are divisible by 5?How many 5 digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.
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Number of ways of forming $8$-digit odd number using only the digits $0, 0, 2, 2, 3, 3, 4, 5$
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Smallest value of n to form 900 n-digit numbers using given digitsFind the sum of all 4 digit numbers which are formed by the digits 1,2,5,6?How many $3$-digit numbers can be formed so that the sum of two digits will be equal to the third digit?Eight digit number is formed using all the digits: $1,1,2,2,3,3,4,5$How many numbers of $7$ digits can be formed with the digit $0,1,1,5,6,6,6$.How many odd numbers of $5$ digits can be formed with the digits $0,2,3,4,5$ without repetition of any digit?Sum of all 5 digit numbers using the digits 1,2,3,4,5 at most once.How many $5$-digit positive even numbers can be formed by using all of the digits $1$, $2$, $3$ and $6$?How many 4 digit numbers can be formed from digits 0 to 9 without repetition which are divisible by 5?How many 5 digit numbers can be formed using digits 1,2,3 with exactly one digit repeating 3 times.
$begingroup$
Let $lambda$ be the number of all possible $8$ digit odd numbers
formed by using only digits $0,0,2,2,3,3,4,5,$ Then $dfraclambda900$ is
My Try: Number is odd if last digit (unit position ) is odd.
So total number of ways of choosing the units place is $dfrac3!2!=3$
Now arranging the extreme left position is $dfrac6!2!cdot 2!$
and arranging all digits is $dfrac6!2!$ ways.
But it have seems that I have done the problem incorrectly.
Could some help me to solve it? Thanks
combinatorics permutations
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
$endgroup$
add a comment |
$begingroup$
Let $lambda$ be the number of all possible $8$ digit odd numbers
formed by using only digits $0,0,2,2,3,3,4,5,$ Then $dfraclambda900$ is
My Try: Number is odd if last digit (unit position ) is odd.
So total number of ways of choosing the units place is $dfrac3!2!=3$
Now arranging the extreme left position is $dfrac6!2!cdot 2!$
and arranging all digits is $dfrac6!2!$ ways.
But it have seems that I have done the problem incorrectly.
Could some help me to solve it? Thanks
combinatorics permutations
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
$endgroup$
add a comment |
$begingroup$
Let $lambda$ be the number of all possible $8$ digit odd numbers
formed by using only digits $0,0,2,2,3,3,4,5,$ Then $dfraclambda900$ is
My Try: Number is odd if last digit (unit position ) is odd.
So total number of ways of choosing the units place is $dfrac3!2!=3$
Now arranging the extreme left position is $dfrac6!2!cdot 2!$
and arranging all digits is $dfrac6!2!$ ways.
But it have seems that I have done the problem incorrectly.
Could some help me to solve it? Thanks
combinatorics permutations
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
$endgroup$
Let $lambda$ be the number of all possible $8$ digit odd numbers
formed by using only digits $0,0,2,2,3,3,4,5,$ Then $dfraclambda900$ is
My Try: Number is odd if last digit (unit position ) is odd.
So total number of ways of choosing the units place is $dfrac3!2!=3$
Now arranging the extreme left position is $dfrac6!2!cdot 2!$
and arranging all digits is $dfrac6!2!$ ways.
But it have seems that I have done the problem incorrectly.
Could some help me to solve it? Thanks
combinatorics permutations
combinatorics permutations
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
edited Mar 31 at 15:33
N. F. Taussig
45.2k103358
45.2k103358
asked Mar 31 at 14:58
DXTDXT
5,8742733
asked Mar 31 at 14:58
DXTDXT
5,8742733
asked Mar 31 at 14:58
DXTDXT
5,8742733
5,8742733
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Another method.
Case 1 (last digit $3$). There are $frac7!2!2!=1260$ numbers in total, in particular, with the first digit $0$. There are $frac6!2!=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.
Case 2 (last digit $5$). There are $frac7!2!2!2!=630$ numbers, in particular, with the first digit $0$. There are $frac6!2!2!=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.
Adding the two cases: $900+450=1350$.
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
$endgroup$
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
add a comment |
$begingroup$
Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.
Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).
Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.
answered Mar 31 at 15:04
kccukccu
11.2k11231
$endgroup$
add a comment |
$begingroup$
Strategy:
Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.
Since the number must be odd, its last digit must be a $3$ or $5$.
If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.
If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.
The number of possible eight-digit odd numbers that can be formed with the given digits is $$binom62binom523! + binom62binom52binom32$$
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
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$begingroup$
Another method.
Case 1 (last digit $3$). There are $frac7!2!2!=1260$ numbers in total, in particular, with the first digit $0$. There are $frac6!2!=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.
Case 2 (last digit $5$). There are $frac7!2!2!2!=630$ numbers, in particular, with the first digit $0$. There are $frac6!2!2!=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.
Adding the two cases: $900+450=1350$.
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
$endgroup$
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
add a comment |
$begingroup$
Another method.
Case 1 (last digit $3$). There are $frac7!2!2!=1260$ numbers in total, in particular, with the first digit $0$. There are $frac6!2!=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.
Case 2 (last digit $5$). There are $frac7!2!2!2!=630$ numbers, in particular, with the first digit $0$. There are $frac6!2!2!=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.
Adding the two cases: $900+450=1350$.
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
$endgroup$
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
add a comment |
$begingroup$
Another method.
Case 1 (last digit $3$). There are $frac7!2!2!=1260$ numbers in total, in particular, with the first digit $0$. There are $frac6!2!=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.
Case 2 (last digit $5$). There are $frac7!2!2!2!=630$ numbers, in particular, with the first digit $0$. There are $frac6!2!2!=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.
Adding the two cases: $900+450=1350$.
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
$endgroup$
Another method.
Case 1 (last digit $3$). There are $frac7!2!2!=1260$ numbers in total, in particular, with the first digit $0$. There are $frac6!2!=360$ numbers with the first $0$ and last $3$. Hence: $1260-360=900$.
Case 2 (last digit $5$). There are $frac7!2!2!2!=630$ numbers, in particular, with the first digit $0$. There are $frac6!2!2!=180$ numbers with the first $0$ and last $5$. Hence: $630-180=450$.
Adding the two cases: $900+450=1350$.
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
answered Mar 31 at 16:44
farruhotafarruhota
22.1k2942
22.1k2942
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
add a comment |
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
It is a nice method, but the wording is incorrect. You meant to say including those with the first digit $0$ rather than with the first digit $0$.
$endgroup$
– N. F. Taussig
Mar 31 at 18:27
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
$begingroup$
@N.F.Taussig, yes, I meant “in particular” in the sense of “including”. And the key word “those” must be added. Let your remark be the errata. Thank you.
$endgroup$
– farruhota
Mar 31 at 18:41
add a comment |
$begingroup$
Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.
Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).
Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.
answered Mar 31 at 15:04
kccukccu
11.2k11231
$endgroup$
add a comment |
$begingroup$
Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.
Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).
Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.
answered Mar 31 at 15:04
kccukccu
11.2k11231
$endgroup$
add a comment |
$begingroup$
Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.
Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).
Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.
answered Mar 31 at 15:04
kccukccu
11.2k11231
$endgroup$
Why do you have factorials when choosing the units digit and leading digit? There is no arrangement happening.
Furthermore you will have to split into a couple cases. You might try to count the number of ways to choose the units digit (it's either $3$ or $5$, so $2$ choices), then count the number of ways to choose the leading digit (it's anything but $0$, so $4$ choices). However, some of these choices are incompatible. E.g., if I choose $5$ for the units digit, it cannot be the leading digit (but choosing $3$ for the units digit and the leading digit is possible).
Beyond this, the number of ways to arrange the remaining digits depends on the digits chosen in the first two steps. For instance, if I choose $4******5$, then I have to arrange $0,0,2,2,3,3$, but if I choose $2******3$ then I have to arrange $0,0,2,3,4,5$. The number of arrangements are different in these two cases.
answered Mar 31 at 15:04
kccukccu
11.2k11231
answered Mar 31 at 15:04
kccukccu
11.2k11231
answered Mar 31 at 15:04
kccukccu
11.2k11231
answered Mar 31 at 15:04
kccukccu
11.2k11231
11.2k11231
add a comment |
add a comment |
$begingroup$
Strategy:
Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.
Since the number must be odd, its last digit must be a $3$ or $5$.
If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.
If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.
The number of possible eight-digit odd numbers that can be formed with the given digits is $$binom62binom523! + binom62binom52binom32$$
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
$endgroup$
add a comment |
$begingroup$
Strategy:
Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.
Since the number must be odd, its last digit must be a $3$ or $5$.
If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.
If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.
The number of possible eight-digit odd numbers that can be formed with the given digits is $$binom62binom523! + binom62binom52binom32$$
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
$endgroup$
add a comment |
$begingroup$
Strategy:
Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.
Since the number must be odd, its last digit must be a $3$ or $5$.
If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.
If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.
The number of possible eight-digit odd numbers that can be formed with the given digits is $$binom62binom523! + binom62binom52binom32$$
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
$endgroup$
Strategy:
Since we wish to form eight-digit odd numbers, the two zeros may not be placed in the first or last positions. Choose two of the middle six positions for the zeros.
Since the number must be odd, its last digit must be a $3$ or $5$.
If the units digit is a $3$, then the remaining five positions must be filled with two $2$s, one $3$, one $4$, and one $5$. Choose two of the five positions for the $2$s. Arrange the three remaining distinct letters in the remaining three positions.
If the units digit is a $5$, then the five remaining positions must be filled with two $2$s, two $3$s, and one $4$. Choose two of the five positions for the $2$s. Choose two of the remaining three positions for the $3$s. The $4$ must be placed in the remaining position.
The number of possible eight-digit odd numbers that can be formed with the given digits is $$binom62binom523! + binom62binom52binom32$$
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
answered Mar 31 at 15:30
N. F. TaussigN. F. Taussig
45.2k103358
45.2k103358
add a comment |
add a comment |
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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
