Question

In how many ways can 11 identical books on English and 9 identical books on Maths be
placed in a row on a shelf so that two books on Maths may not be together?​

Answer 1

Answer:

11 books can be placed in a row in 11! ways.

There will be 12 gaps in which 9 math books can be placed in 12!/(12-9)!

Therefore number of ways in which two math books are not placed together = 11!×12!/3!

if you consider placing 11 english books 1

then ans is 12!/3!

Answer 2

220 ways can 11 identical books on English and 9 identical books on Mathematics be placed in a row on a shelf so that two books on Mathematics may not be together

Given,

English = 11

Mathematics = 9

To find,

number of ways to arrange books

Solution,

If we arrange 11 English books we would get 10gaps in them

*E*E*E*E*E*E*E*E*E*E*E*

Since, here we can see 12 gaps for Mathematics books to fill, we can use combination formula.

12C9 = $\frac{12!}{9![12-9]!} = \frac{(10)(11)(12)}{3!} = (10)(11)(2) = 220.$

Therefore, there are 220 ways in which books can be arranged.