The number of integers between 100 and 1000 s | vedclass

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(D) Let the three-digit number be $N = 100a + 10b + c$,where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.We need to find the number o

Vedclass · Accepted Answer

$70$

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(D) Let the three-digit number be $N = 100a + 10b + c$,where $a \in \{1, 2, \dots, 9\}$ and $b, c \in \{0, 1, \dots, 9\}$.We need to find the number of solutions to $a + b + c = 14$ under the given constraints.Using the stars and bars method with inclusion-exclusion:Let $a' = a - 1$,so $a' \geq 0$. Then $(a' + 1) + b + c = 14 \implies a' + b + c = 13$,where $0 \leq a' \leq 8$,$0 \leq b \leq 9$,and $0 \leq c \leq 9$.The total number of non-negative integer solutions to $a' + b + c = 13$ is $\binom{13+3-1}{3-1} = \binom{15}{2} = 105$.Now,subtract cases where variables exceed their limits:Case $1$: $a' \geq 9$. Let $a'' = a' - 9$,then $a'' + b + c = 4$. Number of solutions = $\binom{4+3-1}{2} = \binom{6}{2} = 15$.Case $2$: $b \geq 10$. Let $b' = b - 10$,then $a' + b' + c = 3$. Number of solutions = $\binom{3+3-1}{2} = \binom{5}{2} = 10$.Case $3$: $c \geq 10$. Let $c' = c - 10$,then $a' + b + c' = 3$. Number of solutions = $\binom{3+3-1}{2} = \binom{5}{2} = 10$.Total valid solutions = $105 - (15 + 10 + 10) = 105 - 35 = 70$.

The number of integers between $100$ and $1000$ such that the sum of their digits is equal to $14$ is:

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