JEE Main 2019 — Functions Question with Solution
From: JEE Main 2019 (Online) 11th January Evening Slot
Question
The number of functions f from {1, 2, 3, ...., 20} onto {1, 2, 3, ...., 20} such that f(k) is a multiple of 3,
whenever k is a multiple of 4, is :
Choose an option
Show full solutionCorrect option: C
Correct answer
C(15)! 6!
Step-by-step explanation
Given that is a multiple of 3 whenever is a multiple of 4, we need to consider how to map elements from the domain {1, 2, 3, ..., 20} to the codomain {1, 2, 3, ..., 20} following this rule.
1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are 5 elements in this subset.
2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.
3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.
4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in ways.
So, combining these two cases, the total number of onto functions is , which corresponds to option C.
1. We first consider the subset of the domain that consists of multiples of 4: {4, 8, 12, 16, 20}. There are 5 elements in this subset.
2. We then consider the subset of the codomain that consists of multiples of 3: {3, 6, 9, 12, 15, 18}. There are 6 elements in this subset.
3. According to the given condition, each of the 5 multiples of 4 must be mapped to a multiple of 3. This can be done in ways, considering that there are 6 options for each of the 5 multiples of 4 (each choice constitutes a combination), and we then consider the permutations of these 5 choices.
4. The remaining 15 elements in the domain (20 original elements minus the 5 multiples of 4) can be mapped onto the remaining 15 elements in the codomain (20 original elements minus the 6 multiples of 3, plus one multiple of 3 that has been assigned to a multiple of 4). This can be done in ways.
So, combining these two cases, the total number of onto functions is , which corresponds to option C.
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