Question

The number of ways that 5 Marathi, 3 English and 3 Tamil books be arranged if the books of each language are to be kept together is

• A. 25920
• B. 9250
• C. 5920
• D. 7480

Answer 1

The Correct Option is A

To solve the problem of arranging the books such that each language's books stay together, we begin by considering each language group as a single "block" or "unit". This results in three blocks: Marathi, English, and Tamil.

First, determine how many ways we can arrange these blocks among themselves. Since there are 3 blocks, they can be arranged in \(3!\) ways, which calculates as follows:

\(3! = 3 \times 2 \times 1 = 6\)

Next, arrange the books within each block. The number of ways to arrange the books within a block is given by the factorial of the number of books in that block:

1. For the 5 Marathi books, they can be arranged in \(5!\) ways: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

2. For the 3 English books, they can be arranged in \(3!\) ways: \(3! = 3 \times 2 \times 1 = 6\)

3. For the 3 Tamil books, they can also be arranged in \(3!\) ways: \(3! = 3 \times 2 \times 1 = 6\)

Thus, the total number of arrangements is the product of the ways to arrange the blocks and the ways to arrange the books within each block:

\(3! \times 5! \times 3! \times 3! = 6 \times 120 \times 6 \times 6\)

Calculating this gives:

\(6 \times 120 = 720\)

\(720 \times 6 = 4320\)

\(4320 \times 6 = 25920\)

Thus, the total number of ways to arrange the books, keeping each language's books together, is 25920.

Answer 2

We can also check this by using the fact that the final count must be a multiple of the internal arrangements within each language group, and testing which option fits.

1. Option A (25920): This must be divisible by \( 5! \times 3! \times 3! = 120 \times 6 \times 6 = 4320 \) since the books within each language can always be shuffled among themselves. \( 25920 \div 4320 = 6 \), which is exactly \( 3! \), the ways to order the three language blocks. This fits perfectly.
2. Option B (9250): \( 9250 \div 4320 \) is not a whole number, so this cannot come from arranging three blocks internally and then in order.
3. Option C (5920): \( 5920 \div 4320 \) is not a whole number either, so this is ruled out.
4. Option D (7480): \( 7480 \div 4320 \) is not a whole number, so this is ruled out too.

Only option A splits cleanly into \( 4320 \times 6 \), matching the block-arrangement structure of the problem.

Let's summarize:

• Any valid count here must be a multiple of \( 5! \times 3! \times 3! = 4320 \).
• Only 25920 divides evenly by 4320, giving the block-order factor of 6.

Therefore, the correct answer is A (25920).