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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Permutation and Combination

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Permutation and Combination

The number of integers, between 100 and 1000 having the sum of...

The number of integers, between 100 and 1000 having the sum of their digits equals to 14 , is _________

Correct answer

0

Step-by-step solution

N=abc\mathrm{N}=\mathrm{ab} \mathrm{c} (i)  All distinct digits a+b+c=14a≥1 b,c∈{0 to  9}\begin{aligned} & \text{ All distinct digits } \\ & \mathrm{a}+\mathrm{b}+\mathrm{c}=14 \\ & \mathrm{a} \geq 1 \\ & \mathrm{~b}, \mathrm{c} \in\{0 \text{\text{ to } } 9\} \end{aligned}byhittrial:8cases(6,5,3)(8,6,0)(9,4,1)(7,6,1)(8,5,1)(9,3,2)(7,5,2)(8,4,2)(7,4,3)(9,5,0)\begin{array}{lll} by hit & trial : & 8 cases \\ (6,5,3) & (8,6,0) & (9,4,1) \\ (7,6,1) & (8,5,1) & (9,3,2) \\ (7,5,2) & (8,4,2) & \\ (7,4,3) & (9,5,0) & \end{array} (ii) 2 same, 1 diffa=b\quad a=b;c 22\mathrm{a}+\mathrm{c}=14byvalues:by values :(3,8)(4,6)(5,4)(6,2)(7,0)}3!2!×5−1=14 cases \begin{aligned} & \left.\begin{array}{l} (3,8) \\ (4,6) \\ (5,4) \\ (6,2) \\ (7,0) \end{array}\right\} \frac{3 !}{2 !} \times 5-1 \\ & =14 \text{ cases } \end{aligned}(iii)allsame:(iii) all same :3a=14a=143× rejected 0 cases \begin{aligned} & \begin{array}{l} 3 a=14 \\ a=\frac{14}{3} \times \text{ rejected } \\ 0 \text{ cases } \end{array} \end{aligned}Hence,Totalcases:Hence, Total cases:8×3!+2×(4)+14=48+22=70\begin{aligned} & 8 \times 3 !+2 \times(4)+14 \\ & =48+22 \\ & =70 \end{aligned}$
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