Mixing
Number of $4$-digit numbers that can be formed with the digits $0,1,2,3,4,5$ with constraint on repetition
$begingroup$
Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated in the same number. How many of these numbers have at least one digit repeated?
For those who wanna skip all the babble below this para on why I thought what I thought just jump to the Modified Problem Statement in the Edit.
Seems pretty easy, right? Until you are me who misread the given constraint on the repetition of the digits as this - "the numbers can be repeated as many times as their magnitude". So, this would imply $0$ being repeated $0$ times, so it will occur only $1$ time. Similarly, $1$ will repeat $1$ time so it will occur only $2$ times, and so $2$ occurs $3$ times, $3$ occurs $4$ times, and so on. If we were to write the multiset for the digits from which we can choose from then, the multiset would be
$$({(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)})$$
And after spending a lot of time I thought why not try to find a solution here, because I was not able to get one.
My Attempt at a solution to my proposed problem:-
The first digit can be anyone of the five non-zero digits. Assuming the non zero digits to be distinct, we get the total no of nonzero digits to be $20$, hence no of ways of selecting the first digit (i.e. the digit at the thousandths place) is $displaystyle{{20}choose{1}}$ and the remaining $20$ digits can be arranged in $3$ remaining spaces is ${}^{20}P_3$. Since we had assumed all the digits to be distinct (even those which were repeating), hence we need to set the overcount that we did due to repetition of arrangements hence we divide the ways of arranging by $(4!)^3.3!.2!.1!$ due to $3,4,5$ can have $4$ copies each in the arrangement, the digit $2$ has $3$ copies and the digit $1$ has $2$ copies. Hence, the total no of ways can be given as
$$dfrac{displaystyle{{20}choose{1}}.{}^{20}P_3}{(4!)^3.3!.2!}$$.
On calculating this comes out to be a fraction. So, this is definitely not the correct approach.
Edit:- For all those who are solving the problem as given in the Problem Statement in the quotes, what I was wanting was that what if instead of the constraint the question had placed the constraint had been what I had interpreted it initially which I have already written just below the quote, i.e. "the numbers can be repeated as many times as their magnitude"
So, lets state the problem that I am trying to solve if you wanna skip all the babble in the first part
Modified Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated as many times as their magnitudes. How many of these numbers have at least one digit repeated?
combinatorics permutations
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
$endgroup$
add a comment |
$begingroup$
Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated in the same number. How many of these numbers have at least one digit repeated?
For those who wanna skip all the babble below this para on why I thought what I thought just jump to the Modified Problem Statement in the Edit.
Seems pretty easy, right? Until you are me who misread the given constraint on the repetition of the digits as this - "the numbers can be repeated as many times as their magnitude". So, this would imply $0$ being repeated $0$ times, so it will occur only $1$ time. Similarly, $1$ will repeat $1$ time so it will occur only $2$ times, and so $2$ occurs $3$ times, $3$ occurs $4$ times, and so on. If we were to write the multiset for the digits from which we can choose from then, the multiset would be
$$({(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)})$$
And after spending a lot of time I thought why not try to find a solution here, because I was not able to get one.
My Attempt at a solution to my proposed problem:-
The first digit can be anyone of the five non-zero digits. Assuming the non zero digits to be distinct, we get the total no of nonzero digits to be $20$, hence no of ways of selecting the first digit (i.e. the digit at the thousandths place) is $displaystyle{{20}choose{1}}$ and the remaining $20$ digits can be arranged in $3$ remaining spaces is ${}^{20}P_3$. Since we had assumed all the digits to be distinct (even those which were repeating), hence we need to set the overcount that we did due to repetition of arrangements hence we divide the ways of arranging by $(4!)^3.3!.2!.1!$ due to $3,4,5$ can have $4$ copies each in the arrangement, the digit $2$ has $3$ copies and the digit $1$ has $2$ copies. Hence, the total no of ways can be given as
$$dfrac{displaystyle{{20}choose{1}}.{}^{20}P_3}{(4!)^3.3!.2!}$$.
On calculating this comes out to be a fraction. So, this is definitely not the correct approach.
Edit:- For all those who are solving the problem as given in the Problem Statement in the quotes, what I was wanting was that what if instead of the constraint the question had placed the constraint had been what I had interpreted it initially which I have already written just below the quote, i.e. "the numbers can be repeated as many times as their magnitude"
So, lets state the problem that I am trying to solve if you wanna skip all the babble in the first part
Modified Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated as many times as their magnitudes. How many of these numbers have at least one digit repeated?
combinatorics permutations
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
$endgroup$
$begingroup$
"will repeat" or can repeat"???
$endgroup$
– barak manos
Dec 6 '16 at 7:39
$begingroup$
"digits can be repeated in the same (4 digit) number" is how I read it.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 7:43
$begingroup$
@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
$endgroup$
– user350331
Dec 6 '16 at 7:48
$begingroup$
As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 8:09
$begingroup$
Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
$endgroup$
– user350331
Dec 6 '16 at 8:13
add a comment |
$begingroup$
Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated in the same number. How many of these numbers have at least one digit repeated?
For those who wanna skip all the babble below this para on why I thought what I thought just jump to the Modified Problem Statement in the Edit.
Seems pretty easy, right? Until you are me who misread the given constraint on the repetition of the digits as this - "the numbers can be repeated as many times as their magnitude". So, this would imply $0$ being repeated $0$ times, so it will occur only $1$ time. Similarly, $1$ will repeat $1$ time so it will occur only $2$ times, and so $2$ occurs $3$ times, $3$ occurs $4$ times, and so on. If we were to write the multiset for the digits from which we can choose from then, the multiset would be
$$({(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)})$$
And after spending a lot of time I thought why not try to find a solution here, because I was not able to get one.
My Attempt at a solution to my proposed problem:-
The first digit can be anyone of the five non-zero digits. Assuming the non zero digits to be distinct, we get the total no of nonzero digits to be $20$, hence no of ways of selecting the first digit (i.e. the digit at the thousandths place) is $displaystyle{{20}choose{1}}$ and the remaining $20$ digits can be arranged in $3$ remaining spaces is ${}^{20}P_3$. Since we had assumed all the digits to be distinct (even those which were repeating), hence we need to set the overcount that we did due to repetition of arrangements hence we divide the ways of arranging by $(4!)^3.3!.2!.1!$ due to $3,4,5$ can have $4$ copies each in the arrangement, the digit $2$ has $3$ copies and the digit $1$ has $2$ copies. Hence, the total no of ways can be given as
$$dfrac{displaystyle{{20}choose{1}}.{}^{20}P_3}{(4!)^3.3!.2!}$$.
On calculating this comes out to be a fraction. So, this is definitely not the correct approach.
Edit:- For all those who are solving the problem as given in the Problem Statement in the quotes, what I was wanting was that what if instead of the constraint the question had placed the constraint had been what I had interpreted it initially which I have already written just below the quote, i.e. "the numbers can be repeated as many times as their magnitude"
So, lets state the problem that I am trying to solve if you wanna skip all the babble in the first part
Modified Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated as many times as their magnitudes. How many of these numbers have at least one digit repeated?
combinatorics permutations
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
$endgroup$
Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated in the same number. How many of these numbers have at least one digit repeated?
For those who wanna skip all the babble below this para on why I thought what I thought just jump to the Modified Problem Statement in the Edit.
Seems pretty easy, right? Until you are me who misread the given constraint on the repetition of the digits as this - "the numbers can be repeated as many times as their magnitude". So, this would imply $0$ being repeated $0$ times, so it will occur only $1$ time. Similarly, $1$ will repeat $1$ time so it will occur only $2$ times, and so $2$ occurs $3$ times, $3$ occurs $4$ times, and so on. If we were to write the multiset for the digits from which we can choose from then, the multiset would be
$$({(0,1),(1,2),(2,3),(3,4),(4,5),(5,6)})$$
And after spending a lot of time I thought why not try to find a solution here, because I was not able to get one.
My Attempt at a solution to my proposed problem:-
The first digit can be anyone of the five non-zero digits. Assuming the non zero digits to be distinct, we get the total no of nonzero digits to be $20$, hence no of ways of selecting the first digit (i.e. the digit at the thousandths place) is $displaystyle{{20}choose{1}}$ and the remaining $20$ digits can be arranged in $3$ remaining spaces is ${}^{20}P_3$. Since we had assumed all the digits to be distinct (even those which were repeating), hence we need to set the overcount that we did due to repetition of arrangements hence we divide the ways of arranging by $(4!)^3.3!.2!.1!$ due to $3,4,5$ can have $4$ copies each in the arrangement, the digit $2$ has $3$ copies and the digit $1$ has $2$ copies. Hence, the total no of ways can be given as
$$dfrac{displaystyle{{20}choose{1}}.{}^{20}P_3}{(4!)^3.3!.2!}$$.
On calculating this comes out to be a fraction. So, this is definitely not the correct approach.
Edit:- For all those who are solving the problem as given in the Problem Statement in the quotes, what I was wanting was that what if instead of the constraint the question had placed the constraint had been what I had interpreted it initially which I have already written just below the quote, i.e. "the numbers can be repeated as many times as their magnitude"
So, lets state the problem that I am trying to solve if you wanna skip all the babble in the first part
Modified Problem Statement:-
Find the number of numbers of four digits that can be made from the digits $0,1,2,3,4,5$ if digits can be repeated as many times as their magnitudes. How many of these numbers have at least one digit repeated?
combinatorics permutations
combinatorics permutations
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
edited Dec 6 '16 at 8:02
user350331
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
asked Dec 6 '16 at 7:27
user350331user350331
1,29021232
1,29021232
$begingroup$
"will repeat" or can repeat"???
$endgroup$
– barak manos
Dec 6 '16 at 7:39
$begingroup$
"digits can be repeated in the same (4 digit) number" is how I read it.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 7:43
$begingroup$
@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
$endgroup$
– user350331
Dec 6 '16 at 7:48
$begingroup$
As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 8:09
$begingroup$
Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
$endgroup$
– user350331
Dec 6 '16 at 8:13
add a comment |
$begingroup$
"will repeat" or can repeat"???
$endgroup$
– barak manos
Dec 6 '16 at 7:39
$begingroup$
"digits can be repeated in the same (4 digit) number" is how I read it.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 7:43
$begingroup$
@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
$endgroup$
– user350331
Dec 6 '16 at 7:48
$begingroup$
As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 8:09
$begingroup$
Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
$endgroup$
– user350331
Dec 6 '16 at 8:13
$begingroup$
"will repeat" or can repeat"???
$endgroup$
– barak manos
Dec 6 '16 at 7:39
$begingroup$
"will repeat" or can repeat"???
$endgroup$
– barak manos
Dec 6 '16 at 7:39
$begingroup$
"digits can be repeated in the same (4 digit) number" is how I read it.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 7:43
$begingroup$
"digits can be repeated in the same (4 digit) number" is how I read it.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 7:43
$begingroup$
@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
$endgroup$
– user350331
Dec 6 '16 at 7:48
$begingroup$
@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
$endgroup$
– user350331
Dec 6 '16 at 7:48
$begingroup$
As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 8:09
$begingroup$
As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
$endgroup$
– Henno Brandsma
Dec 6 '16 at 8:09
$begingroup$
Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
$endgroup$
– user350331
Dec 6 '16 at 8:13
$begingroup$
Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
$endgroup$
– user350331
Dec 6 '16 at 8:13
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5times6times6times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5times5times4times3$$
So the number of them with at least one repeat is $A-B$.
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
$endgroup$
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
add a comment |
$begingroup$
If I read your problem statement correctly, there are no limits on repetitions. As $0$ cannot occur as the first one (by convention), ther are 5 options for the first digit and 6 for the 3 others. So there are $5cdot 6^3 = 1080$ numbers in total. To count the ones with a repeating digit, count those with no repeating digits and take the complement. The first place has $5$ options, the second $5$ as well (all but the one we pick first), then $4$ options and finally $3$, so $25 cdot 12 = 300$ numbers have no repeated digits, so $1080 - 300 = 780$ 4-digit numbers will have at least a repeated digit. I see (in the exact quote you gave) no evidence that 5 can repeat 5 times etc.
In the alternate interpretation, the universe of allowed numbers is smaller: we can have no repeated $0$'s (there are $binom{3}{2} cdot 5 cdot 5$ numbers that are disallowed by that rule (2 out of 3 places with possible 0's, the remaining can be 5 things for the first place and 5 for the remaining digit as well).
As to 1: 2 1's is allowed, but not 3 or 4, so we only discard $binom{4}{3} cdot 5 + 1$ many numbers. The 1 is for $1111$, also disallowed ) After recomputing the size of the "universe" of allowed 4-digit numbers, we still substract all numbers where all digits are different to get the required answer. 4 and 5 offer no restrictions, as we cannot even have more than 4 repeated 4's etc.
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
oldest
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oldest
votes
$begingroup$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5times6times6times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5times5times4times3$$
So the number of them with at least one repeat is $A-B$.
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
$endgroup$
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
add a comment |
$begingroup$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5times6times6times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5times5times4times3$$
So the number of them with at least one repeat is $A-B$.
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
$endgroup$
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
add a comment |
$begingroup$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5times6times6times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5times5times4times3$$
So the number of them with at least one repeat is $A-B$.
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
$endgroup$
The first part is simple. except zero, all digits can go at first and the remaining positions can be either of the six digits. So the total number of 4-digit numbers is:
$$A=5times6times6times6$$
For the second part, the number of 4-digit numbers with no repeat is
$$B=5times5times4times3$$
So the number of them with at least one repeat is $A-B$.
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
edited Dec 6 '16 at 7:52
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
answered Dec 6 '16 at 7:42
msmmsm
6,2292829
6,2292829
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
add a comment |
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
add a comment |
$begingroup$
If I read your problem statement correctly, there are no limits on repetitions. As $0$ cannot occur as the first one (by convention), ther are 5 options for the first digit and 6 for the 3 others. So there are $5cdot 6^3 = 1080$ numbers in total. To count the ones with a repeating digit, count those with no repeating digits and take the complement. The first place has $5$ options, the second $5$ as well (all but the one we pick first), then $4$ options and finally $3$, so $25 cdot 12 = 300$ numbers have no repeated digits, so $1080 - 300 = 780$ 4-digit numbers will have at least a repeated digit. I see (in the exact quote you gave) no evidence that 5 can repeat 5 times etc.
In the alternate interpretation, the universe of allowed numbers is smaller: we can have no repeated $0$'s (there are $binom{3}{2} cdot 5 cdot 5$ numbers that are disallowed by that rule (2 out of 3 places with possible 0's, the remaining can be 5 things for the first place and 5 for the remaining digit as well).
As to 1: 2 1's is allowed, but not 3 or 4, so we only discard $binom{4}{3} cdot 5 + 1$ many numbers. The 1 is for $1111$, also disallowed ) After recomputing the size of the "universe" of allowed 4-digit numbers, we still substract all numbers where all digits are different to get the required answer. 4 and 5 offer no restrictions, as we cannot even have more than 4 repeated 4's etc.
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
add a comment |
$begingroup$
If I read your problem statement correctly, there are no limits on repetitions. As $0$ cannot occur as the first one (by convention), ther are 5 options for the first digit and 6 for the 3 others. So there are $5cdot 6^3 = 1080$ numbers in total. To count the ones with a repeating digit, count those with no repeating digits and take the complement. The first place has $5$ options, the second $5$ as well (all but the one we pick first), then $4$ options and finally $3$, so $25 cdot 12 = 300$ numbers have no repeated digits, so $1080 - 300 = 780$ 4-digit numbers will have at least a repeated digit. I see (in the exact quote you gave) no evidence that 5 can repeat 5 times etc.
In the alternate interpretation, the universe of allowed numbers is smaller: we can have no repeated $0$'s (there are $binom{3}{2} cdot 5 cdot 5$ numbers that are disallowed by that rule (2 out of 3 places with possible 0's, the remaining can be 5 things for the first place and 5 for the remaining digit as well).
As to 1: 2 1's is allowed, but not 3 or 4, so we only discard $binom{4}{3} cdot 5 + 1$ many numbers. The 1 is for $1111$, also disallowed ) After recomputing the size of the "universe" of allowed 4-digit numbers, we still substract all numbers where all digits are different to get the required answer. 4 and 5 offer no restrictions, as we cannot even have more than 4 repeated 4's etc.
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
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$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
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– user350331
Dec 6 '16 at 7:50
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you might wanna see the edit.
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– user350331
Dec 6 '16 at 8:00
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Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
add a comment |
$begingroup$
If I read your problem statement correctly, there are no limits on repetitions. As $0$ cannot occur as the first one (by convention), ther are 5 options for the first digit and 6 for the 3 others. So there are $5cdot 6^3 = 1080$ numbers in total. To count the ones with a repeating digit, count those with no repeating digits and take the complement. The first place has $5$ options, the second $5$ as well (all but the one we pick first), then $4$ options and finally $3$, so $25 cdot 12 = 300$ numbers have no repeated digits, so $1080 - 300 = 780$ 4-digit numbers will have at least a repeated digit. I see (in the exact quote you gave) no evidence that 5 can repeat 5 times etc.
In the alternate interpretation, the universe of allowed numbers is smaller: we can have no repeated $0$'s (there are $binom{3}{2} cdot 5 cdot 5$ numbers that are disallowed by that rule (2 out of 3 places with possible 0's, the remaining can be 5 things for the first place and 5 for the remaining digit as well).
As to 1: 2 1's is allowed, but not 3 or 4, so we only discard $binom{4}{3} cdot 5 + 1$ many numbers. The 1 is for $1111$, also disallowed ) After recomputing the size of the "universe" of allowed 4-digit numbers, we still substract all numbers where all digits are different to get the required answer. 4 and 5 offer no restrictions, as we cannot even have more than 4 repeated 4's etc.
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
$endgroup$
If I read your problem statement correctly, there are no limits on repetitions. As $0$ cannot occur as the first one (by convention), ther are 5 options for the first digit and 6 for the 3 others. So there are $5cdot 6^3 = 1080$ numbers in total. To count the ones with a repeating digit, count those with no repeating digits and take the complement. The first place has $5$ options, the second $5$ as well (all but the one we pick first), then $4$ options and finally $3$, so $25 cdot 12 = 300$ numbers have no repeated digits, so $1080 - 300 = 780$ 4-digit numbers will have at least a repeated digit. I see (in the exact quote you gave) no evidence that 5 can repeat 5 times etc.
In the alternate interpretation, the universe of allowed numbers is smaller: we can have no repeated $0$'s (there are $binom{3}{2} cdot 5 cdot 5$ numbers that are disallowed by that rule (2 out of 3 places with possible 0's, the remaining can be 5 things for the first place and 5 for the remaining digit as well).
As to 1: 2 1's is allowed, but not 3 or 4, so we only discard $binom{4}{3} cdot 5 + 1$ many numbers. The 1 is for $1111$, also disallowed ) After recomputing the size of the "universe" of allowed 4-digit numbers, we still substract all numbers where all digits are different to get the required answer. 4 and 5 offer no restrictions, as we cannot even have more than 4 repeated 4's etc.
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
edited Dec 6 '16 at 8:07
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
answered Dec 6 '16 at 7:42
Henno BrandsmaHenno Brandsma
110k348118
110k348118
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
add a comment |
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
I did write below the quote that I misread the question and was concerned about the method to solve it considering what I had interpreted of the question initially.
$endgroup$
– user350331
Dec 6 '16 at 7:50
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
you might wanna see the edit.
$endgroup$
– user350331
Dec 6 '16 at 8:00
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
$begingroup$
Can you elaborate on how you got the number of numbers which need to be excluded in case of $1$ and $0$. Also, why did you exclude the case where all digits are different.
$endgroup$
– user350331
Dec 6 '16 at 10:10
add a comment |
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"will repeat" or can repeat"???
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– barak manos
Dec 6 '16 at 7:39
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"digits can be repeated in the same (4 digit) number" is how I read it.
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– Henno Brandsma
Dec 6 '16 at 7:43
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@HennoBrandsma - I did solve it correctly as the question asked for, but I was concerned about the way of solving the question the way I interpreted it initially.
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– user350331
Dec 6 '16 at 7:48
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As to your interpretation, I'd then say there can be at most one 1, two 2's etc.
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– Henno Brandsma
Dec 6 '16 at 8:09
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Yeah, I also thought so too, but that would make $0$ occur zero times, but as it is meant to be used, hence provided in the problem as the digits that can be used, so I interpreted it the way I did it. Though it seems rather wrong, well after coming to know of the correct question and my faulty interpretation it seems like it is me putting just a totally different question which just tried to use the words of the original question to just frame my own.
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– user350331
Dec 6 '16 at 8:13