To expand upon what I already said in the comments, the probability of an event, $E$, occurring in an unbiased sample space, $S$, is given by the formula $$Pr(E) = \frac{|E|}{|S|}$$
Note this only applies when every outcome in the sample space is equally likely to occur.
For this problem, our sample space is the set of ways in which the letters in the word ARRANGEMENTS can be arranged. From earlier example, we know that the number of arrangements of a word with $n_a$ copies of $A$, $n_b$ copies of $B$, ..., $n_k$ copies of $K$ with $n=n_a+n_b+\dots+n_k$ letters total will be given by the formula $$\binom{n}{n_a,n_b,\dots,n_k} = \frac{n!}{n_a!n_b!\cdots n_k!}$$
For our specific problem, we know then that ARRANGEMENTS has two $A$'s, two $E$'s, one $G$, one $M$, two $N$'s, two $R$'s, one $S$, and one $T$, and twelve letters total so $|S|=\frac{12!}{2!2!1!1!2!2!1!1!}$ which simplifies as $=\frac{12!}{2!2!2!2!}$ or even further as $=\frac{12!}{(2!)^4}$. (Further simplifications are of course possible, but I will leave it like this so that it is clear how we found the numbers in the first place)
In part (a), it asks us what the probability is that having picked an arrangement of the word ARRANGEMENTS uniformly at random from all possible arrangements, that it has both $E$'s at the front of the word. To do this, we ask the question of how many arrangements of the word ARRANGEMENTS actually satisfies this property.
To count this, let us first guarantee the placement of the two $E$'s at the front. After this, we will place one of the possible arrangements of the remaining ten letters after the $E$'s. Answer the related question of how many ways we can arrange the remaining ten letters. Well, there are two $A$'s, one $G$, one $M$, two $N$'s, two $R$'s, one $S$, and one $T$ with ten letters total to arrange and place after the $E$'s for a total number of such arrangements as $\frac{10!}{(2!)^3}$.
That was the count of how many there are. To get the probability, divide by the sample space size. $$Pr(E)=\frac{|E|}{|S|}=\dfrac{(\frac{10!}{(2!)^3})}{(\frac{12!}{(2!)^4})}=\frac{10!2!}{12!}=\frac{2}{12\cdot 11}=\frac{1}{66}$$
For part (b), we first count how many ways there are to arrange the letters in the word ARRANGEMENTS such that all consonants are together. To do this approach via multiplication principle.
To remind you, the multiplication principle states that if we want to count how many ways we can complete a task and we can describe every outcome uniquely via a sequence of steps with a specific number of options at each step which doesn't rely on the choices that precede it, the total number of outcomes will be the product of the number of options for each step.
- Step one: Arrange the consonants by themselves. This can be accomplished in $\binom{8}{1,1,2,2,1,1}=\frac{8!}{(2!)^2}$ number of ways.
- Step two: Treating the group of consonants that have been arranged in step one as one big piece, lets call it $\mathbb{X}$, arrange the vowels and $\mathbb{X}$ together. We will arrange then $AAEE\mathbb{X}$. This can be accomplished in $\binom{5}{2,2,1}=\frac{5!}{(2!)^2}$ number of ways.
Applying multiplication principle then, there are a total of:
$\frac{8!}{(2!)^2}\cdot \frac{5!}{(2!)^2}=\frac{8!5!}{(2!)^4}$ ways to do this
Applying the definition of probability in an unbiased sample space then, the probability is:
$\dfrac{(\frac{8!5!}{(2!)^4})}{(\frac{12!}{(2!)^4})} = \frac{8!5!}{12!}=\frac{5\cdot 4\cdot 3\cdot 2}{12\cdot 11\cdot 10\cdot 9} = \frac{1}{99}$
Best Answer
Yes, these are all correct - standard approaches to solving these types of problems.