calliiistar
Answered

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2) How many distinguishable letter arrangements can be made from the letters in:
a) FLUKE?
b) PROPOSE?

Sagot :

celumusakhumbulani

Answer:

for "FLUKE" its 5! = 120

and "PROPOSE" its 7! /2!×2! = 1260

Step-by-step explanation:

"FLUKE" has no repeating letters.

"PROPOSE" has two repeating letters.

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

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