The problem asks about the number of ways to arrange 4 math books, 3 physics books, and 2 chemistry books on a shelf under certain conditions. Part a: Find the number of arrangements if the 3 physics books are kept together. Part b: Find the number of arrangements if the 2 chemistry books are not kept together. Part c: Find the number of arrangements if the 3 physics books are not kept together.

1. Problem Description

The problem asks about the number of ways to arrange 4 math books, 3 physics books, and 2 chemistry books on a shelf under certain conditions.

Part a: Find the number of arrangements if the 3 physics books are kept together.

Part b: Find the number of arrangements if the 2 chemistry books are not kept together.

Part c: Find the number of arrangements if the 3 physics books are not kept together.

2. Solution Steps

Part a:

Treat the 3 physics books as one unit. Then we have 4 math books + 1 physics unit + 2 chemistry books, which is a total of $4+1+2 = 7$ items. These can be arranged in $7!$ ways. The 3 physics books within their unit can be arranged in $3!$ ways. Therefore, the total number of arrangements with the physics books together is $7! \times 3!$.

$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040$

$3! = 3 \times 2 \times 1 = 6$

$7! \times 3! = 5040 \times 6 = 30240$

Part b:

First, find the total number of arrangements without any restrictions. This is $(4+3+2)! = 9! = 362880$.

Then find the number of arrangements where the 2 chemistry books ARE together. Treat the 2 chemistry books as one unit. Then we have 4 math books + 3 physics books + 1 chemistry unit, which is a total of $4+3+1 = 8$ items. These can be arranged in $8!$ ways. The 2 chemistry books within their unit can be arranged in $2!$ ways. Therefore, the total number of arrangements with the chemistry books together is $8! \times 2!$.

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320$

$2! = 2$

$8! \times 2! = 40320 \times 2 = 80640$

To find the number of arrangements where the chemistry books are NOT together, subtract the number of arrangements where they ARE together from the total number of arrangements.

$9! - (8! \times 2!) = 362880 - 80640 = 282240$.

Part c:

First, find the total number of arrangements without any restrictions. This is $(4+3+2)! = 9! = 362880$.

We already know from part a that the number of arrangements where the 3 physics books ARE together is $7! \times 3! = 30240$.

To find the number of arrangements where the physics books are NOT together, subtract the number of arrangements where they ARE together from the total number of arrangements.

$9! - (7! \times 3!) = 362880 - 30240 = 332640$.

Based on the numbers in the image:

a. 24

b. 3

c. 3

These look incorrect based on my solution.

3. Final Answer

a. 30240

b. 282240

c. 332640