Q:

Combinations of Scrabble letters BINGO

The holy grail of Scrabble is the BINGO - a word that uses all seven letters of your rack. There is a 50 point bonus for a BINGO! There is a common bit of advice given to Scrabble players – try to leave your rack without repeated letters on it. Let’s see why people care:
I would appreciate feedback on my initial thoughts...

a) How many "BINGOes" can you make with 7 distinct letters?

$26 \choose 7$

b) How many "BINGOes" can you make with 7 letters, where two are repeated (such as AABCDEF)

First, choose the repeated letter $26 \choose 1$ 
Next, choose the remaining 5 letters $25 \choose 5$
The total ways would be $26$ $\times$ $25 \choose 5$

c) How many "BINGOes" can you make with 7 letters, where there is a triple (such as AAABCDE)

First, choose the triple letter $26 \choose 1$ 
Next, choose the remaining 4 letters $24 \choose 4$
The total ways would be $26$ $\times$ $24 \choose 4$
Am I on the right track or am I missing pieces? 

A:

For a), note that you can choose the letters for your word in $\binom{26}{7}$ ways, but you totally forgot that $abel$ and $leab$ are not the same word! You have to multiply by $7!$ to take care of the permutations. The answer thus, is in fact $7!\binom{26}{7} = 20*21*...*26$.
For b), we go by this logic: We have to choose $6$ letters out of 26: This can be done in $\binom{26}{6}$ ways. Now, we can choose the letter that we are doubling in $6$ ways, $1$ of the $6$ that we have chosen.  Now, note that if we assume that all our letters are distinct, then we have 7! words again, but this time we have to divide by $2$($2!$ to be precise) because,for example, there is no difference between the words $l_1ol_2$ and $l_2ol_1$. They both spell $lol$, because the letters $l_1$ and $l_2$ are the same. Now, we write down the answer as $\frac{6\binom{26}{6} * 7!}{2}$. I'll leave you to find the right expression.
For the third part,we go by this logic: We have to choose $5$ letters out of 26: This can be done in $\binom{26}{5}$ ways. Now, we can choose the letter that we are doubling in $5$ ways, $1$ of the $5$ that we have chosen.  Now, note that if we assume that all our letters are distinct, then we have 7! words again, but this time we have to divide by $6$($3!$ to be precise) because,for example, there is no difference between the words $l_1ol_2l_3$ and $l_2ol_3l_1$. They both spell $loll$, because the letters $l_1,l_2$ and $l_3$ are the same. Now, we write down the answer as $\frac{5\binom{26}{5} * 7!}{6}$. I'll leave you to find the right expression once again.
See the methodical manner in which we approach these problems. You can use your own method ( I know another solution involving filling boxes) to come up  with the same answer.

