Question

In how many ways can 4 Latin and 1 English books be placed on a shelf so that the English book is always in the middle, the selection made from 7 Latin and 3 English books?

Answer 1

We will first write down the pattern in which we need our books to be arranged. Then going in one by one we will calculate the number of ways of options with books for each of the places and then add all of them together.

Complete step-by-step solution:
We need to arrange 5 books in total among which 4 are Latin books and 1 is an English book. We have chosen 4 Latin books from the given 7 Latin books and 1 English book from the given 3 English books.
We need an English book in the middle. So, in 5 books, the middle will be the third book.
We need the arrangement to be like Latin Latin English Latin Latin
Now, we will look at its place by place. So, in the first place, we have a Latin book. We have a total of 7 books with us. We can place any one of those 7 in the first place. So, first place has 7 options. Now, we are left with 6 Latin books and 3 English books because 1 Latin has been put into the first place on the shelf. Now, similarly, the second place will be occupied by any of the 6 Latin books. So, second place has 6 options. Now, coming to the third place which has to be taken by any one of 3 English books. So, it will have 3 options. Now, we are left with 5 Latin books and 2 English books, which we are not going to require any further. Now, the fourth place will have 5 options and the last place that is the fifth place will have 4 options.
So, the total number of arrangements will be,
$\Rightarrow$ Total arrangements $= 7 \times 6 \times 3 \times 5 \times 4$
Multiply the terms,
$\Rightarrow$ Total arrangements $= 2520$
Hence, the total number of arrangements is 2520 ways.

Note: The fundamental principle of counting is used to find the total ways.
The fundamental counting principle is a rule used to count the total number of possible outcomes in a situation. It states that if there are $n$ ways of doing something, and $m$ ways of doing another thing after that, then there are $n \times m$ ways to perform both of these actions. In other words, when choosing an option for $n$ and an option for $m$, there are $n \times m$ different ways to do both actions.