SOLUTION: The questions are as follows: On a shelf there are 4 different mathematics books and 8 different English books. i) The books are to be arranged so that the mathematics books

Question 1187402: The questions are as follows:
On a shelf there are 4 different mathematics books and 8 different English books.
i) The books are to be arranged so that the mathematics books are together. In how many different ways can this be done?
Answer: 4!*9! = 8 709 120
I did not have any problem with the question and answer above, but the problem is the next one:
ii) What is the probability that none of the mathematics books are next to each other?
The working given for this question is ((8!)*(9P4))/(12!), but could anyone explain what is the logic behind this working?
Answer by ikleyn(52835) (Show Source): You can put this solution on YOUR website!
.

So, I do not touch part (1) and am focusing on part (2), merely.


Imagine 8 different English books on the shelf.


Assume that they are not tightly placed each to other, having gaps between them.


We should place Math books into these gaps:  one (and no more than one) Math book into each of 7 gaps.


Add to it one potential place on the left of 8 English books and one potential place on the right of 8 English books.


So, you have 7+1+1 = 9 potential places for 4 Math books.


It explains this factor  , which you see in the formula.


Also, 8 English books can be ordered in 8! ways. 

It explains the factor 8! in the formula.



Now, the numerator  8!*C[9]^4  is the number of all favorable dispositions.

The denominator 12! = (8+4)!  is the number of ALL potential placements of the books on the shelf.


The ratio of    to  ,  ,  is the probability, which you are seeking for.

Do you understand my explanation/explanations in full ?

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