How many ways are there to choose 4 letters out of the following 8 letters without duplicates? There are eight letters: $A BCD EF AG$
How many ways can I choose four letters without duplicates?
(1) Eliminating the duplicate letter $A$, no of ways is ${7 \choose 4} = 35$
(2) No of ways to choose $4$ out of $8$ letters is ${8 \choose 4} = 70$
$\quad $ No of duplicates is ${(8-2) \choose 2} = 15$
$\quad $  No of ways is $70 - 15 = 55$
Both seem right, I don't know why and which one of them is wrong?
Thank you.
 A: First way should be the correct one. Note that in your second way, when we say $8 \choose 4$, we consider two $A$'s different. But normally, total number of ways of choosing $4$ out of $8$ letters with duplicates should be
$${6 \choose 4}+{6 \choose 3}+{6 \choose 2} = 50$$
where ${6 \choose 4}$ is the number of choices where we don't choose any $A$'s, ${6 \choose 3}$ is the number of choices where we choose one $A$ and ${6 \choose 2}$ is the number of choices where we choose two $A$'s. So when we exclude the cases where we choose two $A$'s, we get the same answer as in your first way, $35$.
A: See the answer is in question itself. The question says 'duplicates' so the two A's are identical. And if we have to choose different letters, we can eliminate the duplicates and find the no. of ways just like you did in the first method.
A: Alternative approach.

ABCDEFAG

Imagine each element is a building block.  So normally, there are $\binom{8}{4}$ ways of choosing 4 of the 8 blocks.
When you outlaw having more than 1 A, the physical effect is equivalent to taking the two A blocks and gluing them together, so that you can't use them separately.  Now you only have 7 blocks, so the number of ways is $\binom{7}{4}.$
