How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels? How many sequences of length 6 are formed from the 26 letters without repetition where the first or last letter (possibly both) must not be vowels?
I am so lost and confused, but here's my approach:


*

*Total number of sequences should be: 26 * 25 * 24 * 23 * 22 * 21

*First letter not a vowel: 21 * 25 * 24 * 23 * 22 * 21

*Last letter not a vowel:

*

*All 5 vowels used previously: 5 * 4 * 3 * 2 * 1 * 21

*4 vowels are used previously: 5 * 4 * 3 * 2 * 21 * 20

*3 vowels are used previously: 5 * 4 * 3 * 21 * 20 * 19

*2 vowels are used previously: 5 * 4 * 21 * 20 * 19 * 18

*1 vowel is used previously: 5 * 21 * 20 * 19 * 18 * 17



I don't know if this is the right approach and if it is, I don't know where to go from here. Could you please help me understand the methodology of solving these types of problems? 
The solution is given to be: 2x21x5xP(24,4) + 21x20xP(24,4) or else P(26,6) - 5x4xP(24,4) 
 A: The first part of the given solution is a result of:
1)  Counting the number of arrangements with exactly one vowel - either at the beginning or the end:
     a)  Choose whether the vowel goes at the beginning or end:  2 ways.
     b)  Choose a non-vowel to fill the non-vowel spot at the beginning or end: 21 ways.
     c)  Choose a vowel to fill the vowel spot at the beginning or end:  5 ways.
     d)  Choose the sequence of letters to fill the middle P(24,4) ways.
2)  Counting the number of arrangements with no vowels:
     a)  Choose a non-vowel for the first spot:  21 ways.
     b)  Choose a non-vowel for the last spot:  20 ways.
     c)  Choose the sequence of letters to fill the middle P(24,4) ways.
So the total number of acceptable arrangements is  2*21*5*P(24,4) + 21*20*P(24,4).
The "or else" part of the given solution is a result of:
1)  Counting the number of possible sequences of length 6 from the 26 letters without repetition:  P(26,6)  (this is the same as your "total number of sequences")
2)  Subtracting the sequences where both the first and last letters are vowels:
     a)  Choose the two vowels.  This can be done in 5*4 ways.
     b)  Choose the remaining letters.  This can be done in P(24,4) ways.
     c)  The total number of these arrangements is  5*4*P(24,4)
So the total number of acceptable arrangements is  P(26, 6) - 5*4*P(24,4).
I'm out of time for the moment, but if possible, I'll come back and comment about how these compare with your calculations.
A: Note: I'm interpreting the clause "where the first or last letter (possibly both) must not be vowels" as meaning that vowels are forbidden at the slots number 1 and 6. It seems that the intended meaning was: "the set of words whose first and last letters are not both vowels".
You have $26-5=21$ choices for the first letter and $20$ choices for the last letter. The four letters in between can be taken from the $26-2=24$ letters not used so far. This can be done in $24\cdot23\cdot22\cdot21$ ways. The total number $N$ of admissible words is therefore given by
$$N=21\cdot20\cdot24\cdot23\cdot22\cdot21=107110080\ .$$
