Question

2) How many distinguishable letter arrangements can be made from the letters in:

a) FLUKE?
b) PROPOSE?

Answer 1

for "FLUKE" its 5! = 120

and "PROPOSE" its 7! /2!×2! = 1260

Explanation:

"FLUKE" has no repeating letters.

"PROPOSE" has two repeating letters.

we divide by the repeating letters like I did on "PROPOSE" .

THANK YOU.