Question

There are 8 books to be arranged on a shelf. The three novels must be kept together, as do the two math books. The other two books are biology but need not be kept together. How many ways can they be arranged?

Answer 1

96

Explanation:

We have 3 different novels that must be kept together. The number of ways to arrange the 3 novels is `3! = 6`.

We also have 2 different math books that we'll also keep together. The number of ways to arrange the 2 math books is `2! = 2`.

Now to arrange the books on the shelf.

There are 4 places we can place the group of novels (spots `1-3, 2-4, 3-5, 4-6`).

When the novels are in the end spots, `1-3 and 4-6`, there are 2 places where the math books can go (eg. `1-3=>4-5, 5-6`) but when they are in one of the middle spots, they can only be in one place (eg. `2-4=>5-6`). So we need to express this in our calculation. I think the easiest way to do this is to sum up the ways to arrange the groups:

`(("Novel placement","Math placement (ways)"),(1-3,2),(2-4,1),(3-5,1),(4-6,2))`

And so in total there are 8 ways to place the groups of novels, math books, and biology books.

In total then, we have:

`6xx2xx8=96`