Problem 31 - Algebra for JEE ADVANCED

Question

A library has ' \(a\) ' copies of one book, 'b' copies each of two books, \({ }^{2} c\) ' copies each of three books, an single copy of ' \(d\) ' books. The total number of ways in which these books can be arranged in a shelf is equal to a. \(\frac{(a+2 b+3 c+d) !}{a !(b !)^{2}(c !)^{3}}\) b. \(\frac{(a+2 b+3 c+d) !}{a !(2 b !)(c !)^{3}}\) c. \(\frac{(a+b+3 c+d) !}{(c !)^{3}}\) d. \(\frac{(a+2 b+3 c+d) !}{a !(2 b) !(3 c) !}\)

Step-by-Step Solution

Answer

The correct answer is option (a).

Step 1: Total Copies Calculated

First, let's calculate the total number of books. We have 'a' copies of one book, 'b' copies each of two books, 'c' copies each of three books, and 'd' single copies. The total number of books is: \ \( a + 2b + 3c + d \).

Step 2: Understanding Arrangement

The total arrangements of books can be computed with the factorial of the total number of books, \((a + 2b + 3c + d)!\). However, since some books are indistinguishable, we must divide by the factorial of the number of copies for each book type to correct for these repetitions.

Step 3: Correcting for Repetitions

We have: - 'a' identical copies of one book: divide by \(a!\).- 'b' identical copies for two books: divide by \((b!)^2\) since each book has b copies.- 'c' identical copies for three books: divide by \((c!)^3\) since each book has c copies.Thus, the arrangement is given by: \[ \frac{(a + 2b + 3c + d)!}{a! (b!)^2 (c!)^3} \]

Step 4: Matching the Correct Option

Compare the formula derived: \[ \frac{(a + 2b + 3c + d)!}{a! (b!)^2 (c!)^3} \]To the given options, option (a) matches exactly with the derived formula. Therefore, option (a) is the correct answer.

Key Concepts

Arrangement of Books

Imagine you have multiple books and a shelf to place them on. The act of arranging them refers to how you position each book side by side on the shelf. When we talk about arranging books in a combinatorial sense, we're looking at every possible lineup of these books.
If you have different copies of books, this arrangement tends to change. If all books were distinct, the arrangement would simply consider each book a unique piece. However, if some books look alike or are indistinguishable, you have to factor in these repetitions. This means that we don't count arrangements that swap identical books as unique.
This combinatorial exercise becomes interesting when you have groups of identical books. For example:

• 'a' copies of a single book: All identical.
• 'b' copies each of two books: Two sets of books, where each set is identical.
• 'c' copies each of three books: Three sets of books, where each set is identical.
• 'd' single copies: Each distinct.

Understanding these groupings helps in calculating the total arrangement possibilities.

Factorial

When it comes to arranging items, factorial is your go-to mathematical concept. A factorial, denoted by the exclamation mark (!), represents the product of all positive integers up to a particular number.
For instance, if you have 5 different books, the number of ways to arrange them is represented as 5!, which equals:
1 x 2 x 3 x 4 x 5 = 120 ways.
In our discussion, the factorial comes into play in two main areas:

• The total number of ways to arrange all books as if they were distinct, which is \((a + 2b + 3c + d)!\).
• The correction factor for repeated items, dividing by \(a!\), \((b!)^2\), and \((c!)^3\) to account for repetitions and avoid over-counting.

Understanding how factorial works prepares you to tackle the permutations of both distinct and indistinguishable objects.

Indistinguishable Objects

In many combinatorial problems, objects being identical have a significant impact on the counting process. An indistinguishable object, like a twin book in a lineup, complicates how we count possible arrangements.
When we have groups of identical objects, such as 'a' copies of a book, the arrangements amongst these identical copies themselves are not important and are not counted separately. Instead, for each group of indistinguishable objects, we use the factorial of its size as a divisor. This corrects the overestimation of arrangements.
Consider 'b' copies of two different books. Here, since there are two separate sets of identical objects, we divide by \((b!)^2\). This ensures permutations of identical objects don't artificially inflate the count.
In our example, addressing indistinguishable objects is crucial to accurately calculating the number of unique arrangements.

Permutation and Combination

Permutations and combinations are fundamental concepts in combinatorics used to count arrangements and selections. Permutations focus on arranging items. While they look at different possible orders of items, combinations focus on selection without regard to order. In the specific context of arranging books on a shelf, permutations are more relevant because the order matters.
In our scenario:

• The permutation of the entire set of books is initially considered as if every book was unique, which is \((a + 2b + 3c + d)!\).
• To adjust for indistinguishable books, we divide by \(a!\) for 'a' identical books, \((b!)^2\) for the two sets of 'b' identical books each, and \((c!)^3\) for the three sets of 'c' identical books each.

These adjustments through divisions help us realize the appropriate permutations by pruning out redundancy from indistinct arrangements, aligning the solving process with real-world configurations on a shelf.