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Find the number of seven-digit positive integers such that the sum of the digits is 19.

$\binom{24}{6} - \binom{15}{6} - 6\binom{14}{6}$

Denote the digits from left to right by $x_1,... , x_7$. Then the answer is the number of solutions of $x_1 + ... + x_7 = 19$ in non-negative integers not exceeding 9, but with the additional restriction that $x_1$ is positive.

This is the question and answer from the book "Math of Choice". My question is why did the author uses 25 instead of 24 since there are 7$x$, hence, shouldn't it be $\binom{19+7-1}{7-1}$?

Kindly advise

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    $\begingroup$ You have to subtract $1$ to cope with the condition that $x_1>0$. $\endgroup$
    – lulu
    May 23, 2022 at 22:34
  • $\begingroup$ @lulu, thank you for the reply. If I were to subtract 1 to account for $x_1 > 0$, shouldn't i be subtracting 1 from 6 too? Which derives to $\binom{24}{5}$ $\endgroup$ May 23, 2022 at 22:45
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    $\begingroup$ If we let $X_1=x_1-1$ the problem is now to count the solutions to $X_1+x_2+\cdots +x_7=18$ in non-negative integers. This is a routine Stars and Bars problem, and the usual formula applies. $\endgroup$
    – lulu
    May 23, 2022 at 22:49

2 Answers 2

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Not quite a routine stars and bars question.

The first digit has to be $\geq 1$, whereas the others can be $\geq 0$

We can equalize the lower limits by substituting $X_1 = x_1-1$ as suggested by @lulu to make the equation
$X_1 +x_2 +...+x_7=18$ in non-negative integers,

but then the upper limit for $X_1$ is $8$, while it is $9$ for the rest, and we need to take care of this idiosyncrasy while applying inclusion-exclusion,

that is why the answer will be $\dbinom{24}{6} - \dbinom11\dbinom{24-9}{6} - \dbinom61\dbinom{24-10}{6}$

= $\dbinom{24}6- \dbinom{15}6 - 6\dbinom{14}6$

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I have a stupid question :

Can I replace $\binom{1}{1}\binom{24-9}{6}$ by $\binom{18-9+5}{5}$? (It means in case $x_1=9$). Thus:

$\binom{24}{6}-\binom{14}{5}-6\binom{14}{6}$

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