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 the

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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 $x \in \{1, 2, \dots, 9\}$ and $y, z \in \{0, 1, \dots, 9\}$.Let $T = x - 1$,so $x = T + 1$. Since $1 \leq x \leq 9$,we have $0 \leq T \leq 8$.Substituting this into the equation: $(T + 1) + y + z = 10 \implies T + y + z = 9$.We need to find the number of non-negative integer solutions to $T + y + z = 9$ subject to the constraints $T \leq 8$,$y \leq 9$,and $z \leq 9$.The total number of non-negative integer solutions without constraints is given by the formula $\binom{n+r-1}{r-1}$,where $n=9$ and $r=3$: $\binom{9+3-1}{3-1} = \binom{11}{2} = \frac{11 	imes 10}{2} = 55$.Now,we subtract the cases that violate the constraints:$1$. If $T = 9$,then $y=0$ and $z=0$. This is $1$ case ($x=10$,which is not a digit).$2$. If $y > 9$ or $z > 9$,there are no such cases since the sum is only $9$.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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