Question

How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d? (c a, b and c are next to each other?

Answer 1

A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
$\text{Case 1: } 4 \cdot 4!$

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
$\text{Case 2: } 3 \cdot 4!$

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
$\text{Case 3: } 2 \cdot 4!$

$\text{Case 4: } 1 \cdot 4!$

$\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240$

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

$\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36$