Jtdylktuy

I tried casework for this. For the 2 Es case, I chose 2 non-Es to exclude then arranged the remaining 6 letters. Please pardon the formatting; I’m on my phone.

n(0 Es)+n(1 E)+n(2 Es)
=6! + 7P6 + (6C2)(6!/2!)

Can someone let me know if my work is right? Thank you!

Almost.

Number of arrangements of six letters with no Es: There are six letters other than E in STAMPEDE. Since the letters are distinct, they can be arranged in $6!$ ways.

Number of arrangements of six letters with exactly one E: We must choose which five of the other six letters will be used with the E. Since the six letters are all distinct, they can be arranged in $6!$ ways. Thus, the number of arrangements with exactly one E is
$$binom{6}{5}6!$$
This where you made your mistake.

Number of arrangements of six letters with two Es: We must choose which four of the other six letters will be used with the Es. Thus, we must arrange six letters, two of which are Es and with each of the others appearing exactly once. We choose two of the six positions for the Es. The remaining four distinct letters can be arranged in the remaining four positions in $4!$ ways. Hence, there are
$$binom{6}{4}binom{6}{2}4! = frac{6!}{4!2!} cdot frac{6!}{2!4!} cdot 4! = binom{6}{2} cdot frac{6!}{2!}$$
as you found.

Total: $$6! + binom{6}{5}6! + binom{6}{4}binom{6}{2}4!$$