Number of $7$-digit positive integers with digital sum equal to $48$?












1












$begingroup$


How many $7$-digit positive integers are there with digit sum $48$? Leading zero is not allowed.



My approach : consider a number $x_1x_2x_3x_4x_5x_6x_7$ where $x_1$ isn't $0$. Now, the solution of $x_1+ dots + x_7=48$ in the set of $W$ is $C(54,6)$ where $x_1$ can be $0$ too, which we don't want and we need to get rid of the solutions starting with $0$. Number of solutions starting with $0$ of this equation : $0+x_2+x_3+ dots +x_7=48$ is $C(53,5)$, which is the number we want to get rid of from $C(54,6)$. Thus required answer would be $C(54,6)-C(53,5)$. But the answer given is way different than my one. If any, where have I went wrong?






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You may have included numbers where the largest digit is bigger than $9$
    $endgroup$
    – Henry
    Dec 26 '18 at 8:52










  • $begingroup$
    I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 8:57






  • 3




    $begingroup$
    Did you exclude invalid cases like $x_1=48$?
    $endgroup$
    – drhab
    Dec 26 '18 at 9:18












  • $begingroup$
    Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
    $endgroup$
    – hardmath
    Dec 26 '18 at 18:08
















1












$begingroup$


How many $7$-digit positive integers are there with digit sum $48$? Leading zero is not allowed.



My approach : consider a number $x_1x_2x_3x_4x_5x_6x_7$ where $x_1$ isn't $0$. Now, the solution of $x_1+ dots + x_7=48$ in the set of $W$ is $C(54,6)$ where $x_1$ can be $0$ too, which we don't want and we need to get rid of the solutions starting with $0$. Number of solutions starting with $0$ of this equation : $0+x_2+x_3+ dots +x_7=48$ is $C(53,5)$, which is the number we want to get rid of from $C(54,6)$. Thus required answer would be $C(54,6)-C(53,5)$. But the answer given is way different than my one. If any, where have I went wrong?






combinatorics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    You may have included numbers where the largest digit is bigger than $9$
    $endgroup$
    – Henry
    Dec 26 '18 at 8:52










  • $begingroup$
    I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 8:57






  • 3




    $begingroup$
    Did you exclude invalid cases like $x_1=48$?
    $endgroup$
    – drhab
    Dec 26 '18 at 9:18












  • $begingroup$
    Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
    $endgroup$
    – hardmath
    Dec 26 '18 at 18:08














1












1








1





$begingroup$


How many $7$-digit positive integers are there with digit sum $48$? Leading zero is not allowed.



My approach : consider a number $x_1x_2x_3x_4x_5x_6x_7$ where $x_1$ isn't $0$. Now, the solution of $x_1+ dots + x_7=48$ in the set of $W$ is $C(54,6)$ where $x_1$ can be $0$ too, which we don't want and we need to get rid of the solutions starting with $0$. Number of solutions starting with $0$ of this equation : $0+x_2+x_3+ dots +x_7=48$ is $C(53,5)$, which is the number we want to get rid of from $C(54,6)$. Thus required answer would be $C(54,6)-C(53,5)$. But the answer given is way different than my one. If any, where have I went wrong?






combinatorics







share|cite|improve this question











$endgroup$




How many $7$-digit positive integers are there with digit sum $48$? Leading zero is not allowed.



My approach : consider a number $x_1x_2x_3x_4x_5x_6x_7$ where $x_1$ isn't $0$. Now, the solution of $x_1+ dots + x_7=48$ in the set of $W$ is $C(54,6)$ where $x_1$ can be $0$ too, which we don't want and we need to get rid of the solutions starting with $0$. Number of solutions starting with $0$ of this equation : $0+x_2+x_3+ dots +x_7=48$ is $C(53,5)$, which is the number we want to get rid of from $C(54,6)$. Thus required answer would be $C(54,6)-C(53,5)$. But the answer given is way different than my one. If any, where have I went wrong?






combinatorics




combinatorics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 10:04









N. F. Taussig

45.2k103358




45.2k103358










asked Dec 26 '18 at 8:48









Epsilon zeroEpsilon zero

34118




34118








  • 1




    $begingroup$
    You may have included numbers where the largest digit is bigger than $9$
    $endgroup$
    – Henry
    Dec 26 '18 at 8:52










  • $begingroup$
    I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 8:57






  • 3




    $begingroup$
    Did you exclude invalid cases like $x_1=48$?
    $endgroup$
    – drhab
    Dec 26 '18 at 9:18












  • $begingroup$
    Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
    $endgroup$
    – hardmath
    Dec 26 '18 at 18:08














  • 1




    $begingroup$
    You may have included numbers where the largest digit is bigger than $9$
    $endgroup$
    – Henry
    Dec 26 '18 at 8:52










  • $begingroup$
    I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 8:57






  • 3




    $begingroup$
    Did you exclude invalid cases like $x_1=48$?
    $endgroup$
    – drhab
    Dec 26 '18 at 9:18












  • $begingroup$
    Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
    $endgroup$
    – hardmath
    Dec 26 '18 at 18:08








1




1




$begingroup$
You may have included numbers where the largest digit is bigger than $9$
$endgroup$
– Henry
Dec 26 '18 at 8:52




$begingroup$
You may have included numbers where the largest digit is bigger than $9$
$endgroup$
– Henry
Dec 26 '18 at 8:52












$begingroup$
I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
$endgroup$
– Epsilon zero
Dec 26 '18 at 8:57




$begingroup$
I have included numbers 0<=xi<=9, 1<=i<=7 for the first one I believe.
$endgroup$
– Epsilon zero
Dec 26 '18 at 8:57




3




3




$begingroup$
Did you exclude invalid cases like $x_1=48$?
$endgroup$
– drhab
Dec 26 '18 at 9:18






$begingroup$
Did you exclude invalid cases like $x_1=48$?
$endgroup$
– drhab
Dec 26 '18 at 9:18














$begingroup$
Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
$endgroup$
– hardmath
Dec 26 '18 at 18:08




$begingroup$
Let me suggest "sum of digits" is more apt than "digital sum", since the latter is often used to signify the iterative process of summing digits to get a single digit (also called "casting out nines").
$endgroup$
– hardmath
Dec 26 '18 at 18:08










1 Answer
1






active

oldest

votes


















1












$begingroup$

Your approach is wrong: $C(54,6)$ assumes that digitis like '32' or '15' are allowed. This is not a "stars and bars" problem.



Denote with $Z(l,s)$ the number of integers of length $l$ with sum of digits equal to $s$ assuming that leading zero is allowed.



You have the following recurrence formula:



$$Z(l,s)=sum_{d=0}^{d=9}Z(l-1,s-d)$$



...with the following exit criteria:



$$Z(1,s)=0 quad text{if} quad s<0 lor s>9$$



$$Z(1,s)=1 quad text{if}quad 0 le s le 9$$



You are actually trying to calculate $Z(7, 48)-Z(6,48)$ (this eliminates all 7-digit numbers with a leading zero):



#include <iostream>

using namespace std;



int count(int length, int sum) {

    if(length == 1) 

        return (sum >= 0 && sum <= 9)? 1: 0;

    int cnt = 0;

    for(int i = 0; i <= 9; i++) 

        cnt += count(length - 1, sum - i);

    return cnt;

}



int main() {

    cout << (count(7, 48) - count(6, 48));

}


...and the result is 50568.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:18












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1 Answer
1






active

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1 Answer
1






active

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active

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active

oldest

votes









1












$begingroup$

Your approach is wrong: $C(54,6)$ assumes that digitis like '32' or '15' are allowed. This is not a "stars and bars" problem.



Denote with $Z(l,s)$ the number of integers of length $l$ with sum of digits equal to $s$ assuming that leading zero is allowed.



You have the following recurrence formula:



$$Z(l,s)=sum_{d=0}^{d=9}Z(l-1,s-d)$$



...with the following exit criteria:



$$Z(1,s)=0 quad text{if} quad s<0 lor s>9$$



$$Z(1,s)=1 quad text{if}quad 0 le s le 9$$



You are actually trying to calculate $Z(7, 48)-Z(6,48)$ (this eliminates all 7-digit numbers with a leading zero):



#include <iostream>

using namespace std;



int count(int length, int sum) {

    if(length == 1) 

        return (sum >= 0 && sum <= 9)? 1: 0;

    int cnt = 0;

    for(int i = 0; i <= 9; i++) 

        cnt += count(length - 1, sum - i);

    return cnt;

}



int main() {

    cout << (count(7, 48) - count(6, 48));

}


...and the result is 50568.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:18
















1












$begingroup$

Your approach is wrong: $C(54,6)$ assumes that digitis like '32' or '15' are allowed. This is not a "stars and bars" problem.



Denote with $Z(l,s)$ the number of integers of length $l$ with sum of digits equal to $s$ assuming that leading zero is allowed.



You have the following recurrence formula:



$$Z(l,s)=sum_{d=0}^{d=9}Z(l-1,s-d)$$



...with the following exit criteria:



$$Z(1,s)=0 quad text{if} quad s<0 lor s>9$$



$$Z(1,s)=1 quad text{if}quad 0 le s le 9$$



You are actually trying to calculate $Z(7, 48)-Z(6,48)$ (this eliminates all 7-digit numbers with a leading zero):



#include <iostream>

using namespace std;



int count(int length, int sum) {

    if(length == 1) 

        return (sum >= 0 && sum <= 9)? 1: 0;

    int cnt = 0;

    for(int i = 0; i <= 9; i++) 

        cnt += count(length - 1, sum - i);

    return cnt;

}



int main() {

    cout << (count(7, 48) - count(6, 48));

}


...and the result is 50568.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:18














1












1








1





$begingroup$

Your approach is wrong: $C(54,6)$ assumes that digitis like '32' or '15' are allowed. This is not a "stars and bars" problem.



Denote with $Z(l,s)$ the number of integers of length $l$ with sum of digits equal to $s$ assuming that leading zero is allowed.



You have the following recurrence formula:



$$Z(l,s)=sum_{d=0}^{d=9}Z(l-1,s-d)$$



...with the following exit criteria:



$$Z(1,s)=0 quad text{if} quad s<0 lor s>9$$



$$Z(1,s)=1 quad text{if}quad 0 le s le 9$$



You are actually trying to calculate $Z(7, 48)-Z(6,48)$ (this eliminates all 7-digit numbers with a leading zero):



#include <iostream>

using namespace std;



int count(int length, int sum) {

    if(length == 1) 

        return (sum >= 0 && sum <= 9)? 1: 0;

    int cnt = 0;

    for(int i = 0; i <= 9; i++) 

        cnt += count(length - 1, sum - i);

    return cnt;

}



int main() {

    cout << (count(7, 48) - count(6, 48));

}


...and the result is 50568.






share|cite|improve this answer









$endgroup$



Your approach is wrong: $C(54,6)$ assumes that digitis like '32' or '15' are allowed. This is not a "stars and bars" problem.



Denote with $Z(l,s)$ the number of integers of length $l$ with sum of digits equal to $s$ assuming that leading zero is allowed.



You have the following recurrence formula:



$$Z(l,s)=sum_{d=0}^{d=9}Z(l-1,s-d)$$



...with the following exit criteria:



$$Z(1,s)=0 quad text{if} quad s<0 lor s>9$$



$$Z(1,s)=1 quad text{if}quad 0 le s le 9$$



You are actually trying to calculate $Z(7, 48)-Z(6,48)$ (this eliminates all 7-digit numbers with a leading zero):



#include <iostream>

using namespace std;



int count(int length, int sum) {

    if(length == 1) 

        return (sum >= 0 && sum <= 9)? 1: 0;

    int cnt = 0;

    for(int i = 0; i <= 9; i++) 

        cnt += count(length - 1, sum - i);

    return cnt;

}



int main() {

    cout << (count(7, 48) - count(6, 48));

}


...and the result is 50568.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 26 '18 at 10:07









OldboyOldboy

9,40711138




9,40711138












  • $begingroup$
    Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:18


















  • $begingroup$
    Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
    $endgroup$
    – Epsilon zero
    Dec 26 '18 at 10:18
















$begingroup$
Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
$endgroup$
– Epsilon zero
Dec 26 '18 at 10:18




$begingroup$
Thanks a lot ! I will look into the stars and bars method again to clarify it's idea.
$endgroup$
– Epsilon zero
Dec 26 '18 at 10:18


















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