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How many ten digit numbers have the sum of their digits equal to $4$? Well, my solution goes like this:

The first digit of the $10$ digit number cannot be $0$. It can be $1,2,3$ or $4$ only . If the first digit is $1$ then we can have $3$ other digits as $1$ so as the sum of digit is $4$ . This can be done in $9\choose3$ ways . Now when the 1st digit is $1$ then we can have two digits $2$ and $1$ as another case such as sum of digit remains $4$. So, this can be done in $9\choose 2$ ways . Now, another case can be the one when 1st digit is $4$ but another digit is $3$ . This can be done in $9\choose 1$ ways . So the number of ways when 1st digit is $1$ is $9\choose3$$+$$9\choose2$$+$$9\choose1$ways. Now, if the 1st digit is $2$ then the other two digits can be $1$ each . This can done in $9\choose2$ways . If the 1st digit is $2$ the other digit can be $2$ . This can be done in $9\choose1$ways.The total of ways this can be done is $9\choose2$$+$$9\choose1$ ways. If the 1st digit is $3$ then another digit among those $9$ digits must be $1$.this can be done in $9\choose1$ways. If the 1st digit is $4$ then all the other digits are zero . This can be done in $1$ way only. So, the total number of ways in which the sum of digits can be $4$ in a $10$ - digit number is $9\choose3$$+$$9\choose2$$+$$9\choose1$$+$$9\choose2$$+$$9\choose1$$+$ $9\choose1$$+$$1$ ways$ =184$ways

However the answer in the book is given as: $1+2$$9\choose 1$$+$$9\choose1$$+$$9\choose2$$3!$$/2!$$+$$9\choose 3$$=$$220$ways

Where is the mistake? Where is the problem occuring?

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    $\begingroup$ The correct answer is $\binom{12}{3} = 220$, which can be demonstrated using the stars and bars method. Your error is when you consider the case where the first digit is 1 and there is also a 2. There are not $\binom{9}{2}$ ways of doing this; there are $P^9_2$ ways, since which digit is a 2 and which a 1 matters. $\endgroup$ Apr 23, 2022 at 4:07
  • $\begingroup$ Related and this too. $\endgroup$
    – user983440
    Apr 23, 2022 at 4:09
  • $\begingroup$ @Yooo yes...but I wanted to verify my solution... $\endgroup$
    – user992622
    Apr 23, 2022 at 4:10
  • $\begingroup$ @MarkSaving Thanks a lot! I do get it now... $\endgroup$
    – user992622
    Apr 23, 2022 at 4:12
  • $\begingroup$ You added incorrectly. The expression you obtained adds to $184$. Since $P(9, 2) = 2\binom{9}{2}$, adding $\binom{9}{2}$ to $184$ does give the correct answer of $220$. $\endgroup$ Apr 23, 2022 at 9:42

1 Answer 1

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As Mark Saving pointed out in the comments, the error you made was not taking into account the order of the digits $1$ and $2$ in numbers in the case in which the leading digit is $1$ and the other nonzero digits are $1$ and $2$. In that case, there are $9$ ways to place the second $1$ and eight ways to place the $2$. With that correction, you would have obtained $220$ rather than $184$ since $P(9, 2) - \binom{9}{2} = 9 \cdot 8 - 36 = 36$.

Here is another approach to the problem. Let $x_i$ be the $i$th digit. Then $$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} = 4 \tag{1}$$ Since the leading digit cannot be zero, the integer $x_1 \geq 1$. Each of the remaining variables represents a nonnegative integer.

Let $x_1' = x_1 - 1$. Then $x_1'$ is also a nonnegative integer. Substituting $x_1' + 1$ for $x_1$ in equation 1 yields \begin{align*} x_1' + 1 + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 4\\ x_1' + x_2 + x_3 + x_4 + x_5 + x_6 + x_7 + x_8 + x_9 + x_{10} & = 3 \tag{2} \end{align*} Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of $10 - 1 = 9$ addition signs in a row of $3$ ones. For instance, $$+ + 1 + + + + 1 1 + + +$$ corresponds to the solution $x_1' = x_2 = 0$, $x_3 = 1$, $x_4 = x_5 = x_6 = 0$, $x_7 = 2$, $x_8 = x_9 = x_{10} = 0$ (in which case, the original number was $10,010,002,000$ since $x_1 = x_1' + 1$). The number of solutions of equation 2 is the number of ways we can insert $10 - 1 = 9$ addition signs in a row of $3$ ones, which is $$\binom{3 + 10 - 1}{10 - 1} = \binom{12}{9} = 220$$ since we must choose which nine of the twelve positions required for three ones and nine addition signs will be filled with addition signs.

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