Meaning of 'there are exactly $2$ letters between any $2$ 'E'' (Permutation and Combination) In a game show, the host gives an incorrect arrangement of the letters of the word 'EXCELLENT' and lets the contestant guess the word within a given time limit. Find the number of different ways that the host can arrange the $9$ letters if 
(i) the first and last letters must be occupied by consonants,
(ii) there are exactly $2$ letters between any $2$ 'E'
The $9$ letters in 'EXCELLENT' are now separated into $3$ groups of $3$ letters each. In how many ways can the letters be grouped so that no group has more than one E and no group has more than one L?
My attempt:
(i) $4\times 3 \times \frac{7!}{3! 2!} + 1 \times 4 \times \frac{7!}{3!} \times 2 + 1 \times \frac{7!}{3!}$ 
(I separate into $4$ cases, where L is not chosen as the first and last letter, first is L while last is not L, first is not L while last is L, and both first and last are L)
(ii) I don't understand the question. Since there are $3$ letters 'E', the first 'E' and the last 'E' must contain more than $2$ letters. Can anyone explain the intention of this part?
For the last part, since there are only $3$ letters 'E', each group must contain exactly one 'E'. 
Assume that first and second groups must contain letters 'E' and 'L' while the last group contains only letter 'E' but not 'L'. 
Then we have $4\times 3! + 3 \times 3! + 3!$. 
Are my answers correct?
 A: 
Find the number of ways the letters of the word EXCELLENT can be arranged if the first and last positions must be occupied by consonants.

Your answer is correct.
As a check, observe that the total number of arrangements of the letters of the word EXCELLENT is 
$$\frac{9!}{3!2!}$$
The only vowels are the E's.  There are 
$$\frac{8!}{2!2!}$$
arrangements that begin with an E.  By symmetry, the same number of arrangements end with an E.  The number of arrangements that both begin and end with an E is 
$$\frac{7!}{2!}$$
By the Inclusion-Exclusion Principle, the number of arrangements that do not begin and end with consonants is 
$$2 \cdot \frac{8!}{2!2!} - \frac{7!}{2!} = \frac{8!}{2!} - \frac{7!}{2!} = \frac{1}{2}(8! - 7!)$$
Hence, the number of arrangements that do begin and end with a consonant is 
$$\frac{9!}{3!2!} - \frac{1}{2}(8! - 7!) = 12600$$
which agrees with your answer.

Find the number of ways the letters of the word EXCELLENT can be arranged if there are exactly two letters between any two E's.  

If there are exactly two letters between any two E's, then there are three possibilities.  The E's are in 


*

*positions $1$, $4$, and $7$ $$E \square \square E \square \square E \square \square$$

*positions $2$, $5$, and $8$ $$\square E \square \square E \square \square E \square$$

*positions $3$, $6$, and $9$ $$\square \square E \square \square E \square \square E$$


In each of the three cases, the remaining six letters can be arranged in 
$$\frac{6!}{2!}$$
ways, so the number of arrangements of the word EXCELLENT in which there are exactly two letters between any two E's is 
$$3 \cdot \frac{6!}{2!}$$
of which one is the word EXCELLENT itself. If the host wishes to give an incorrect arrangement of the letters of the word EXCELLENT in which there are exactly two letters between any two E's, he has 
$$3 \cdot \frac{6!}{2!} - 1$$
options.

The letters of the word EXCELLENT are grouped into three groups of three letters each.  In how many ways can the letters be grouped so that no group has more than one E and no group has more than one L?

The context of this question is not clear.
As you observed, there is an E in each group and an L in two of the three groups.  That leaves us with four distinct consonants.  There are $\binom{4}{2} = 6$ ways of selecting the letters that are placed in the group without an L.  Since each of the two remaining letters is placed in a group with both an E and an L, there are six possible groups if the groups are neither labeled nor ordered.  
If the groups are labeled (say $A$, $B$, and $C$) but not ordered, there are $3! \cdot 6$ ways to arrange the letters in three groups.  
Since each such group has three different letters, each group can be ordered internally in $3!$ ways.  
