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I have $5$ different physics textbooks and $4$ different chemistry textbooks. In how many ways can I place the $9$ textbooks on a bookshelf, in a row, if there must be a chemistry textbook in the middle, and there must be a physics textbook at each end

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redsoxrock200

Answer:

57600 ways

Explanation:

For the place in the middle we have 4 books so the number of combinations for that place is 4.

For the places at the end of the shelf I have 5 books, I need 2, so the number is a combination:

2· \frac{5!}{(5-2)!·2!}=2·10=20

For the 6 places on the shelf I have 6 books so the number is a combination 6!=720

So, the number of combinations is:

4·20·720=57600.

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fichoh

Answer: 57600

Explanation:

Given the following;

5 different physics textbooks

4 different chemistry textbooks

There are 9 different positions for arranging the textbooks on a shelf.

If there must be a physics textbook at both ends;

Number of physics textbooks = 5

Both ends = 2 different positions

Mathematically,

5P2 = 5! ÷ (5-2)!

5! ÷ 3!2! =(5×4)

5P2 = 20 ways

If a chemistry textbook must be in the middle;

Number of chemistry textbooks = 4

Middle = 1 position

Mathematically,

4P1 = 4! ÷ (4-1)!

4P1 = 4! ÷ 3!

4P1 = 4 ways

3 positions have been calculated,

The 6 positions left = 6!

6! = 720ways

Therefore,

6! × 5P2 × 4P1 = 720 × 20 × 4 = 57600