Question

The number of numbers from 1 to ${10}^{100}$ having the sum of their digits equal to 3 is

• A. 171100
• B. 177100
• C. 171700
• D. 100710

Answer 1

The correct option is C

171700

Consider $100$ cells corresponding to $100$ digits. lesser digit numbers will have leading zeroes. for eaxample $3$ is a $1-$digit number represented as $\mathrm{00......003}$ with $99$ leading zeroes. sum of the digits $3$, occurs as a single $3$, or a $2$ and a $1$ or there $1^{\prime }s$.
A single $3$:
there are $100$ numbers
there are ${}^{100}{P}_2=100\times 99=9900$ numbers
three $1^{\prime }s$:
there are ${}^{100}{C}_3=\frac{100\cdot 99\cdot 98}{6}=161700$ numbers
total$=100+9900+161700=171700$