Consider lists of length 4 made with symbols A,B,C,D,E,F,G. How many are there if repetition is allowed and the list has an E? This question was from the Book of Proof by Richard H. I initially thought I can do $(1 \cdot 7^3) \cdot 4 = 1372$, because I thought if one entry has E, then all other entries can have any symbol.
I understood I was wrong when I read the solution:

Now we seek the number of length-$4$ lists where repetition is allowed and the list must contain an E. Here is our strategy: By Part (a) of this exercise there are $7 \cdot 7 \cdot 7 \cdot 7 = 7^4 = 2401$ lists with repetition allowed. Obviously this is not the answer to our current question, for many of these lists contain no E. We will subtract from $2401$ the number of lists that do not contain an E. In making a list that does not contain an E, we have six choices for each list entry (because we can choose any one of the six letters A, B, C, D, F or G). Thus there are $6 \cdot 6 \cdot 6 \cdot 6 = 6^4 = 1296$ lists without an E. So the answer to our question is that there are $2401 − 1296 = 1105$ lists with repetition allowed that contain at least one E.

but I am still not sure why my approach would have $267$ more lists. I guess I don't understand what $(1 \cdot 7^3) \cdot 4$ would represent.
 A: To list the difference between your attempt and the solution, consider the number of strings that have exactly $i$ Es:

*

*Exactly $1$ E: $(1\cdot 6^3)\cdot 4 = 864$

*Exactly $2$ Es: $6^2\cdot \binom42 = 216$

*Exactly $3$ Es: $6^1\cdot \binom43 = 24$

*Exactly $4$ Es: $6^0\cdot \binom 44 = 1$
The given solution sums the above counts:
$$864+216+24+1 = 1105 \text{}$$
While your attempt incorrectly overcounted strings, by counting each string the number of times E appears:
$$864\cdot 1+216\cdot2+24\cdot3+1\cdot 4 = 1372$$
A: It looks like you are attempting to count sequences which contain an E by placing an E, then filling in the remaining letters.  However, if you are going to do that, you need to account for the number of Es and their locations.  Otherwise, you will count cases with more than one E multiple times.
Exactly one E:  There are four ways to place the E.  There are six choices for each of the remaining three positions.  Hence, there are $4 \cdot 6^3$ such sequences.
Exactly two Es:  There are $\binom{4}{2}$ ways to select two of the four positions for the Es.  There are six choices for each of the remaining two positions.  Hence, there are $\binom{4}{2} \cdot 6^2$ such sequences.
Exactly three Es:  There are $\binom{4}{3}$ ways to select three of the four positions for the Es.  There are six choices for the remaining position.  Hence, there are $\binom{4}{3} \cdot 6$ such sequences.
Exactly four Es:  There is one way to fill all four positions with Es.
Total:  Since the four cases above are mutually exclusive and exhaustive, there are
$$\binom{4}{1}6^3 + \binom{4}{2}6^2 + \binom{4}{3}6 + \binom{4}{4} = 1105$$
sequences with at least one E.
Why did you obtain $267$ more lists?
You counted each sequence with two Es twice, each sequence with three Es thrice, and the sequence with four Es four times.
Consider, for example, the sequence ABEE.  You counted this sequence twice, depending on which position you designated as the E that must appear in the sequence: $AB\color{red}{E}E, ABE\color{red}{E}$.
Similarly, you counted the sequence AEEE three times, again depending on which E you designated as the E that must appear in the sequence:  $A\color{red}{E}EE, AE\color{red}{E}E, AEE\color{red}{E}$.
You counted the sequence EEEE four times, depending on which E you designated as the E that must appear in the sequence:  $\color{red}{E}EEE, E\color{red}{E}EE, EE\color{red}{E}E, EEE\color{red}{E}$.
Notice that
$$1 \cdot 4 \cdot 7^3 = \binom{4}{1}6^3 + \color{red}{2}\binom{4}{2}6^2 + \color{red}{3}\binom{4}{3}6 + \color{red}{4}\binom{4}{4}$$
We only want to count each sequence once.  Therefore, the excess you added by designating a particular E as the E that must be in the sequence is
$$\binom{4}{2}6^2 + 2 \cdot \binom{4}{3}6 + 3\binom{4}{4} = 216 + 48 + 1 = 267$$
A: I thought of it slightly differently but if you rearrange its essentially the same method:
$$\text{number of possible combinations}=7^4=2401$$
$$\text{probability of a list not containing E}=\left(\frac67\right)^4$$
$$\text{probability of a list containing E}=1-\left(\frac67\right)^4$$
$$\text{number of combinations containing E}=\text{number of possible combinations}\times \text{probability of a list containing E}=7^4\left[1-\left(\frac67\right)^4\right]=1105$$
Maybe this method will make more sense to you. As others have commented your method would be fine except that there will be many repetitions, such as:
$$AAAE,BBBE,EEEE$$ etc.
A: This is essentially the same as Henry Lee's calculation, but just counting, without bringing in probability. If you temporarily ignore the requirement of having at least one E, the number of possibilities is $7^4=2401$. Now pay attention to the "contain an E" requirement. Among the $2401$ options above, the number of possibilities that don't contain an E is $6^4=1296$. Therefore, the number of possibilities that do contain an E is $2401-1296=1105$.
