Question

The number of integers which lie between $1$ and ${10}^6$ and which have the sum of the digits equal to $12$ is

• A. $8550$
• B. $5382$
• C. $6062$
• D. $8055$

Answer 1

Answer: (C)

$6062$

To find the number of integers between $1$ to ${10}^6$ which have sum of the digits is $12$.

That is, the number between $1$ to $1,000,000$

Now, $1,000,000$ have sum of the digits $1$.

So, we need to find the number of integers between $1$ to $999,999$ which have sum of the digits is $12$

The numbers have maximum $6$ digits.

So, we are trying to find the number of solutions of the equation given below

$d_{!}+d_2+d_3+d_4+d_5+d_6=12,\text{where}0\le d_i\le 9$ ........... $\left(i\right)$

Let's forget about the restriction first and try to find out the number of solutions of $\left(i\right)$ with $d_i\ge 0$

If we think there are $12$ balls and $6$ boxes, and we need to put all these balls(with repetition) in these $6$ boxes,

The total no. of possible ways to do so is ${}^{6+12-1}{C}_{12}={}^{17}{C}_{12}$

Now, considering count of balls in each boxes, we can get the number.

But, the upper bound of $9$ for each digit is not mentioned.

Though this violation of upper bound can only exist in one place, because $10+3>12$

Let's say it has occured at first position.

So, we need to decide number of solutions to the equation

$d_1^{\prime }+d_2+d_3+d_4+d_5+d_6=2$ with $d_i\ge 0$

which is ${}^{6+2-1}{C}_2={}^7{C}_2$

If any one of the six positions get violated, total no. of violation $=6\times {}^7{C}_2$

So, the number of solutions to equation $\left(i\right)$ is ${}^{17}{C}_{12}-6\times {}^7{C}_2$

$=6188-126=6062$

Hence, the answer is $6062$.