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Mathematics · Permutations and Combinations

JEE Main 2025 — 24 January, Evening Shift — Question 19

Number of functions f:{1,2,...,100}→{0,1},f : \{1, 2, ..., 100\} \rightarrow \{0, 1\}, that assign 1 to exactly one of the positive integers less than or equal to 98, is equal to _____\_\_\_\_\_.

Answer: 392

Numerical answer — enter this value.

Step-by-step solution

We   need   the   number   of   functions   f:{1,2,…,100}→{0,1} such   thatexactly   one   of   the   integers   ≤98   is   assigned   the   value   1.Choose   which    element   in {1,2,…,98} gets   f(x)=1:  98   choices.The   remaining   97   elements   get f(x)=0.For   f(99) and   f(100),   each   can   independently   be   0   or   1.∴Number   of   functions  =98×22=98×4=392.392\begin{aligned} &\text{We\; need\; the\; number\; of\; functions\; } f : \{1, 2, \ldots, 100\} \rightarrow \{0, 1\} \text{ such\; that} \\[4pt] &\text{exactly\; one\; of\; the\; integers\; } \leq 98\; \text{ is\; assigned\; the\; value\; } 1. \\[8pt] &\text{Choose\; which \; element\; in } \{1, 2, \ldots, 98\} \text{ gets\; } f(x) = 1: \; 98\; \text{ choices.} \\[6pt] &\text{The\; remaining\; } 97 \; \text{ elements\; get } f(x) = 0. \\[6pt] &\text{For\; } f(99) \text{ and\; } f(100),\; \text{ each\; can\; independently\; be\; } 0\; \text{ or\; } 1. \\[6pt] &\therefore \text{Number\; of\; functions\;} = 98 \times 2^2 = 98 \times 4 = 392. \\[6pt] &\boxed{392} \end{aligned}
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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Applications of Permuations and Combination

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