Injective Functions
How many one-to-one functions can be defined from a set containing 4 distinct elements into a set containing 6 distinct elements, assuming standard function definitions apply?
Select the correct option:
Solution
360
A one-to-one (injective) function assigns distinct images to distinct domain elements, so images must be chosen without repetition, a key idea linking functions to permutations in JEE Advanced. The first domain element has 6 possible images, the second has 5 remaining, the third 4, and the fourth 3, since each must avoid previously used codomain values. Multiplying gives 6×5×4×3 = 360 injective functions, which equals the permutation P(6,4). This sequential-choice argument is the standard archetype for counting injections. Option 1296 = 6^4 counts all functions allowing repetition, violating injectivity. Option 24 = 4! counts only bijections of a 4-set, ignoring the larger codomain. Option 1080 is an arithmetic distractor with no structural meaning. Therefore the number of injective functions is 360. Plausibility check: the count must be less than the total 6^4 = 1296 and greater than 4! = 24, and 360 comfortably lies between, confirming the permutation reasoning is dimensionally sensible.
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- injective functions
- Difficulty
- Easy
- Year
- 2025
Solution
Correct Answer:
360
A one-to-one (injective) function assigns distinct images to distinct domain elements, so images must be chosen without repetition, a key idea linking functions to permutations in JEE Advanced. The first domain element has 6 possible images, the second has 5 remaining, the third 4, and the fourth 3, since each must avoid previously used codomain values. Multiplying gives 6×5×4×3 = 360 injective functions, which equals the permutation P(6,4). This sequential-choice argument is the standard archetype for counting injections. Option 1296 = 6^4 counts all functions allowing repetition, violating injectivity. Option 24 = 4! counts only bijections of a 4-set, ignoring the larger codomain. Option 1080 is an arithmetic distractor with no structural meaning. Therefore the number of injective functions is 360. Plausibility check: the count must be less than the total 6^4 = 1296 and greater than 4! = 24, and 360 comfortably lies between, confirming the permutation reasoning is dimensionally sensible.
This easy difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of injective functions. It appeared in the 2025 exam.
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