The total number of 3-digit numbers,whose sum | vedclass

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(A) Let the $3$-digit number be represented as $xyz$,where $x$ is the hundreds digit,$y$ is the tens digit,and $z$ is the units digit. We are given th

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$54$

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(A) Let the $3$-digit number be represented as $xyz$,where $x$ is the hundreds digit,$y$ is the tens digit,and $z$ is the units digit. We are given the condition $x + y + z = 10$,where $1 \leq x \leq 9$ and $0 \leq y, z \leq 9$. Let $T = x - 1$,so $x = T + 1$. Since $1 \leq x \leq 9$,we have $0 \leq T \leq 8$. Substituting into the equation: $(T + 1) + y + z = 10 \implies T + y + z = 9$. The number of non-negative integer solutions to $T + y + z = 9$ is given by the formula $\binom{n+r-1}{r-1} = \binom{9+3-1}{3-1} = \binom{11}{2} = \frac{11 	imes 10}{2} = 55$. However,we must exclude cases where the digits exceed $9$. Since $T \leq 8$,$y \leq 9$,and $z \leq 9$,the only case to exclude is when $T=9$ (which implies $x=10$,not possible for a digit). If $T=9$,then $y=0$ and $z=0$. This is $1$ case. Thus,the total number of valid $3$-digit numbers is $55 - 1 = 54$.

The total number of $3$-digit numbers,whose sum of digits is $10$,is

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