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A set of 7 distinct books needs to be arranged on a shelf. However, two specific books cannot be placed next to each other. In how many different ways can the books be arranged in the shelf.

 

Option: 1

1250

 


Option: 2

1500

 


Option: 3

3600

 


Option: 4

2000


Answers (1)

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To calculate the number of different ways the 7 distinct books can be arranged on a shelf, with the condition that two specific books cannot be placed next to each other, we can use the concept of permutations.

First, let's consider the total number of arrangements without any restrictions. Since there are 7 distinct books, they can be arranged on the shelf in 7 ! (7 factorial) ways.

Next, let's consider the number of arrangements where the two specific books are placed next to each other. We can treat these two books as a single entity. Thus, we have 6 distinct entities to arrange on the shelf.

Since there are 6 distinct entities, they can be arranged in 6 ! ways.

However, within the arrangement of these 6 entities, the two specific books can be swapped. So, for each arrangement, there are 2 ! ( 2 factorial) ways to arrange the two specific books.

Therefore, the number of arrangements where the two specific books are placed next to each other is 6 ! \times 2 !.

To find the number of arrangements where the two specific books cannot be placed next to each other, we subtract the number of arrangements where the two specific books are placed next to each other from the total number of arrangements:

7 !-(6 ! \times 2 !)

Calculating this expression, we get:

7 !-(6 ! \times 2 !)=5,040-(720 \times 2)=5,040-1,440=3,600 .

Therefore, there are 3,600 different ways to arrange the 7 distinct books on the shelf, with the condition that two specific books cannot be placed next to each other.

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Riya

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