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Distribution Into Distinct Boxes

Easymathematics

The number of ways to place 5 distinct letters into 3 distinct mailboxes, where any mailbox may receive any number of letters including none, equals which value?

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About This Question

Subject
mathematics
Chapter
permutations and combinations
Topic
distribution into distinct boxes
Difficulty
Easy
Year
2025
Tags
advanced-calculus-drilldistributiondistinct-objectsdistinct-boxesmultiplication-principle

Solution

Correct Answer:

Distributing distinct objects into distinct boxes with no restriction gives each object an independent choice of box, an application of the multiplication principle in JEE Advanced. Each of the 5 distinct letters can independently go into any of the 3 mailboxes, providing 3 choices per letter. By the multiplication principle, the total number of distributions is 3^5 = 243. The exponent is the number of objects and the base is the number of boxes, the canonical form (boxes)^(objects). Option 125 = 5^3 reverses base and exponent. Option 15 = 5 × 3 wrongly adds choices. Option 10 = C(5,2) confuses this with a selection. Hence there are 243 ways. Plausibility check: this distinct-into-distinct count equals the number of functions from a 5-set to a 3-set, which is 3^5 = 243, consistent with each letter being assigned exactly one mailbox. A combination equation conceals a quadratic in the unknown count, and spotting the product of consecutive integers usually resolves it faster than formal factoring. These equations recur whenever a geometric or selection count is specified and the underlying population size is sought, so fluency in reversing C(n,2) and C(n,3) values into n is a worthwhile exam reflex.

This easy difficulty mathematics question is from the chapter permutations and combinations, covering the topic of distribution into distinct boxes. It appeared in the 2025 exam.

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