The total number of all those 3-digit numbers | vedclass

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(A) Let the $3$-digit number be represented as $abc$,where $a$ is the hundreds digit,$b$ is the tens digit,and $c$ is the units digit. Given the condi

Vedclass · Accepted Answer

$54$

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(A) Let the $3$-digit number be represented as $abc$,where $a$ is the hundreds digit,$b$ is the tens digit,and $c$ is the units digit. Given the condition $a + b + c = 10$,where $1 \leq a \leq 9$ and $0 \leq b, c \leq 9$. Let $a' = a - 1$,so $a = a' + 1$. Substituting this into the equation: $(a' + 1) + b + c = 10 \Rightarrow a' + b + c = 9$,where $a', b, c \geq 0$. The number of non-negative integral solutions is given by the formula $^{n+r-1}C_{r-1}$,where $n=9$ and $r=3$: $^{9+3-1}C_{3-1} = ^{11}C_2 = \frac{11 	imes 10}{2} = 55$. However,we must exclude cases where any digit exceeds $9$. Since $a = a' + 1$,if $a' = 9$,then $a = 10$,which is not possible for a digit. This corresponds to the solution $(a', b, c) = (9, 0, 0)$,which gives $a = 10, b = 0, c = 0$. Thus,we subtract this $1$ invalid case: $55 - 1 = 54$. Therefore,the total number of such $3$-digit numbers is $54$.

The total number of all those $3$-digit numbers in which the sum of all the digits in each of them is $10$,is

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