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
Taking into account the standard formulas for total permutations and permutations with repetitions, the $11$ letters can be arranged in $11!\,$ ways, of which $$\frac {11!}{2!2!2!}=4.989.600\,\,\, \,$$ are different. Since in any arrangement we can switch two identical letters without changing the letter sequence, and because there are three pairs of equal letters (A, M, and T), this $1:8 \,$ ratio between distinct (i.e., taking into account repetitions) and general (i.e. ignoring repetitions) sequences is also valid for the subset of valid arrangements, that is to say those which satisfy the conditions described in the OP. As usual, the searched probability can be obtained by dividing the number of valid arrangements (distinct or general) to the corresponding number of total ones (again distinct or general). Clearly the resulting ratio is the same, irrespective of whether we consider distinct or general arrangements in both the numerator and the denominator. In this answer, I will focus on the number of distinct arrangements.
To count the distinct valid arrangements, since vowels and consonants must be at the same position, we can start by considering them separately as a sequence of $4$ vowels and another sequence of $7$ consonants. So we can continue as follows. For the vowels, we have to count only the arrangements where the letters A are not in the first or third position in the vowel sequence, i.e. those where they are in the second and fourth position (which also implies that the other two vowels are not in the original position). There are clearly $2$ distinct arrangements of this kind, given by EAIA and IAEA.
For the seven consonants, we can start by placing the two M, reminding that they cannot be in the first or fourth position in the consonant sequence. Let us consider three cases:
1) the two M are in the second and fifth position of the consonants sequence (i.e. in the original places of the T). Then the remaining five letters can be arranged in $5!/2!=60\,\,$ distinct ways. These $ 60$ sequences can have one or more among the letters H, C, S (but not a T) in the original position, so they are not all valid. To select the valid sequences among these, we can note that, choosing any of the three letters HCS (e.g., the C), $1/5$ of the sequences contains it in the original position. This is also true for the other two letters (in this example, H and S): because the groups of permutations in which these three letters are in their original position are overlapped, the proportion of sequences that includes at least one of the three letters in the original position can be obtained as $$3 \cdot \frac{1}{5}-3 \cdot \frac {1}{5 \cdot 4}+\frac {1}{5 \cdot 4 \cdot 3}=\frac {28}{60}$$ giving a total of $32$ valid arrangements.
2) only one of the two M is in the second or fifth position of the consonant sequence and the other is in another position. There are $ 2 \cdot 5 =10 \,\,$ distinct ways of placing the M in this manner. In this case, one of the places originally occupied by the T is still blank. We can fill this place in $3$ ways, choosing one letter among H, C, or S. There remain four places that have to be filled with two T and the two remaining letters among H, C, or S. This can be done in $4!/2! =12$ distinct ways. However, among these $12$ arrangements, some must be excluded because the two remaining letters among H, C, or S cannot be positioned in their original places. By an argument similar to that used in point 1, the proportion of arrangement to be excluded is $$2\cdot \frac {1}{4} -\frac {1}{4 \cdot 3}=\frac {5}{12}$$ All this procedure gives $10 \cdot 3 \cdot 7=210\,\,\,$ valid arrangements.
3) none of the two M is placed in the second or fifth position of the consonant sequence. There are $ \binom {5}{2}=10\,\,\,$ distinct ways of placing the M in this manner. In this way, both places originally occupied by the T are still blank. We can fill these two places in $6$ ways choosing two letters among H, C, or S. There remain three places that have to be filled with two T and the only remaining letter among H, C, or S. This can be done in $3!/2! =3$ distinct ways. However, among these $3$ arrangements, one must be excluded because the only remaining letter among H, C, or S cannot be positioned in its original place. All this procedure gives $10 \cdot 6 \cdot 2=120\,\,\,$ valid arrangements.
So, collecting the results above, there are $32+210+120=362\,\,\,\,\,$ valid arrangements for the consonants. Multiplying it by the $2$ valid arrangements of the vowels, we get $724\,$ distinct valid arrangements. Dividing these by the total number, we obtain a probability of
$$\frac {724}{4.989.600} \approx 0.000145$$
corresponding to about one possibility out of $6892\,$.