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A house number begins with two letters. If the possible letters are A, B, C, D and E, how many different permutations of these letters can be made if no letter is used more than once?

Written by Winona · 18 sec read >


A house number begins with two letters. If the possible letters are A, B, C, D and E, how many different permutations of these letters can be made if no letter is used more than once?

Answer:

20

Step-by-step explanation:

The formula for permutation is  nPr = \frac{n!}{(n-r!)}

n = the total number of objects

r = the total number of objects selected

In your given problem, there are five possible letters in total. n = 5

Each number begins with two letters. r = 2

Substitute the numbers in the formula.

nPr = \frac{5!}{(5-2!)}

The answer will be 20.


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