Question

In how many ways can 4 Latin, and 1 English book 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

Hint: 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 answer:
We need to arrange 5 books in total among which 4 are Latin books and 1 is English book. We have chosen 4 Latin books from the given 7 Latin books and 1 English book from given 3 English books.
We need the English book in the middle. So, in 5 books, the middle will be the third book.
We basically need the arrangement to be like: Latin Latin English Latin Latin
Now, we will look at it place by place. So, in the first place we have Latin book. We have in total 7 books with us. We can place any one of those 7 at 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. Hence, the 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.

Therefore, the total number of arrangements will be $7 \times 6 \times 3 \times 5 \times 4 = 2520$ ways.

Note: The alternate way to solve this is: We require 4 books among the 7 Latin books and 1 book among the 3 English books. Hence, the number of ways will be $^7{C_4}{ \times ^3}{C_1}$.
We know that $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$
Hence, the number of arrangements are $^7{C_4}{ \times ^3}{C_1} = \dfrac{{7 \times 6 \times 5}}{{3 \times 2}} \times \dfrac{3}{1} = 2520$ ways.