Logic using Permutation I have this question in front of me:

How many different four letter word can be formed using the letters of “power” such that at least one letter is repeated within the new word.

I solved this using this concept:
I have four places, so at each place there can be 5 letters placed, assuming 2 places as one (because the letted in both the places is same) and multiplying $5×5×5$ I get 125.
Now because in those 2 places any of the 5 nos. can come so I will multiply 125 by 5 again which gives me 625, but this is wrong... the correct ans is 505.
Kindly help me where I am wrong.
 A: Ignoring the restriction, there are $5^4=625$ possible four-letter words that can be formed. The ones with no repeated letters – the excluded words – number $^5P_4=\frac{5!}{(5-4)!}=120$. Therefore there are $625-120=505$ admissible words.
A: In addition to the economical "negative space" approach given by Parcly Taxel, we can build up the answer by cases.
case $1$: one repeated letter, 3 sub-cases:
case $1a$: 2 copies of repeated letter: choose the repeated letter, place in two spots, chose the remaining letters: $5\cdot \binom 42 \cdot 4\cdot 3  = 5\cdot 6 \cdot 12 = 360$ options.
case $1b$: 3 copies of repeated letter: choose the repeated letter, place in three spots, chose the remaining letter: $5\cdot \binom 43 \cdot 4 = 5\cdot 4\cdot 4 = 80$ options.
case $1c$: 4 copies of repeated letter: choose the repeated letter: $5$ options.
case $2$: two repeated letters, must be two repeats for each. Choose the letter that goes first, choose the other location for that, choose the other letter: $5\cdot \binom 31 \cdot 4 = 5\cdot 3\cdot 4 = 60$ options
Total of $360+80+5+60 = 505$ options as before.

Note that you can get into error on case $2$ if you say "choose a letter, choose two places for it, choose the other letter", because the same configuration will come up twice.
