Combinatorics method by splitting

Question

this is the question:

In a trial examination session a candidate at a school has to take 18 examinations papers including, the physics paper, chemistry paper, and the biology paper. No more than 2 of these 3 papers may be taken consecutively. There is no restrictions on the order in which the other examination papers may be taken. Find the number of different orders in which these 18 examinations papers may be taken.

i'm wondering how you would approach this, i've thought of doing $\overbrace{n(\text{number of possibilities where they are not next to each other})}^{\text{what they ask}}=n(\text{total arrangements of 18 examinations})-n(\text{arrang when bio and physics are together})-n(\text{arrang when bio and chem are together})-n(\text{arrang when physics and bio are together})$

Is this logic in the correct way?

Answer 1

The question asks for no more than 2 our of these 3 paper to be taken consecutively. It means that you could have 2 in a row, but not three in a row, you are deducing too much in your calculation.

One way to solve the problem is

1. For simplicity assign numbers $1,2,\ldots,18$ to each paper. And say that the three specific papers are 1, 2 and 3.
2. We calculate the total number of different orders in which the 18 papers could be taken, whithout taking into consideration the restrictions on paper $1,2,3$.

With no restriction, this is simply $T = 18!$

3. We calculate how many ways there are to take the 18 papers WITH paper 1,2,3 in consecutive order.

There are 6 ways to take these 3 papers $(123,132,213,\ldots)$. For each one, there are 16 position they can be taken at. For example for (123), we could have $(123xxx\ldots, x123xxx\ldots,xx123xxx\ldots, \ldots)$. And for each one of these, there are $15!$ possible orders for the other papers. Therefore the total number of ways to take the exam papers WITH papers 1,2,3 in consecutive order is $$C = 6 \times 16 \times 15!=6 \times 16!$$

4. Finally, your final result is just the difference of the two.

$$N=T-C=18!-6 \times 16! = 300 \times 16!$$