# 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.

• The first way is correct. The second way incorrectly distinguishes between the outcome $\color{red}{A}BCD$ and the outcome $\color{blue}{A}BCD$, treating them as different when in reality they are the same (where the red $A$ originated from the start of your collection and the blue $A$ from the end of your collection as $\color{red}{A}BCDEF\color{blue}{A}G$) Nov 28, 2020 at 14:01

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, $$35$$.
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 $$\binom{7}{4}.$$