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.)
-
A(a + b + c + d)! / (a! b! c!)
-
B(a + 2b + 3c + d)! / {a! (b!)2 (c!)3}
-
C(a + 2b + 3c + d)! / (a! b! c!)
-
DNone of these
Answer
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}