A library has a copies of one title, b copies of each of two titles, c copies of each of three titles, and a single copy of d distinct titles (one each). In how many ways can all these books be arranged in a row? (Indistinguishable copies of the same title.)

Difficulty: Medium

Correct Answer: (a + 2b + 3c + d)! / {a! (b!)2 (c!)3}

Explanation:


Introduction / Context:
This is a multiset permutation problem. We have totals of identical copies by title. The denominator multiplies factorials of each title’s multiplicity.


Given Data / Assumptions:

  • Total books N = a + 2b + 3c + d.
  • Indistinguishable copies for identical titles.
  • Two titles have b copies each ⇒ (b!)^2; three titles have c copies each ⇒ (c!)^3.


Concept / Approach:
Number of distinct arrangements of N items with groups of indistinguishable items is N! divided by the product of factorials of each group size.


Step-by-Step Solution:

Arrangements = (a + 2b + 3c + d)! / [a! * (b!)^2 * (c!)^3 * (1!)^d].Since each of the d single-copy titles has multiplicity 1, their 1! factors are 1 and need not be written.


Verification / Alternative check:
Setting b = c = 0 reduces to (a + d)! / a!, consistent with a identical copies of one title and d singletons.


Why Other Options Are Wrong:
Options lacking (b!)^2 or (c!)^3 fail to account for indistinguishability across multiple titles of equal multiplicity.


Common Pitfalls:
Forgetting that “two books with b copies each” contributes (b!)^2 and “three books with c copies each” contributes (c!)^3.


Final Answer:
(a + 2b + 3c + d)! / {a! (b!)2 (c!)3}

More Questions from Permutation and Combination

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion