Finding the no of ways to count the letters in an English alphabet How many strings of six lower case letters from the English
alphabet contain
a) the letter $a$?
b) the letters $a$ and $b$?
c) the letters $a$ and $b$ in consecutive positions with $a$
preceding $b$, with all the letters distinct?
d) the letters $a$ and $b$, where $a$ is somewhere to the left
of $b$ in the string, with all the letters distinct?
What I tried:
a) to find the no of ways to get letter $a$, I took 
no of ways = 
total no of ways without restriction-does not contain letter $a$
thus no of ways $= 26^6 - 25^6$
b) to find the no of ways to get letters $a$ and $b$, I took 
no of ways = 
total no of ways without restriction-(does not contain letter $a$ + does not contain letter $b$)
thus no of ways = $26^6-25^6-25^6$
I'm unsure of how to do part c) and part d). Could anyone explain? Thanks.
 A: Your answer for the first part is correct.

How many strings of six lower case letters from the English alphabet contain the letters $a$ and $b$?

You subtracted the number of strings that do not contain $a$ and the number of strings that do not contain $b$ from the total number of strings to obtain $26^6 - 2 \cdot 25^6$.  However, you have subtracted the strings that contain neither $a$ nor $b$ twice.  Thus, we have to add those $24^6$ strings to the total.  By the Inclusion-Exclusion Principle, the number of strings that contain both $a$ and $b$ is $$26^6 - 2 \cdot 25^6 + 24^6$$

How many strings of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ in consecutive positions with $a$ preceding $b$.

Think of $ab$ as a block.  Since the letters are distinct, we must select four other letters from the remaining $24$ letters, which we can choose in $\binom{24}{4}$ ways. We now have five objects to arrange, the block $ab$ and the four other letters we just selected.  They can be permuted in $5!$ ways.  Hence, the number of strings of six distinct lower case letters from the English alphabet that contain the letters $a$ and $b$ in consecutive positions with $a$ preceding $b$ is $$\binom{24}{4} \cdot 5!$$

How many strings of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ with $a$ preceding $b$.

Since the letters are distinct, we must choose $4$ of the remaining $24$ letters, which we can do in $\binom{24}{4}$ ways.  We can arrange those six letters in $6!$.  However, in half of those permutations, $b$ precedes $a$.  Thus, the number of six distinct lower case letters from the English alphabet contain the letters $a$ and $b$ with $a$ preceding $b$ is 
$$\frac{1}{2} \cdot \binom{24}{4} \cdot 6!$$
