How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 assuming that repetition of the digits is not allowed?
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Question

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4, and 5 assuming that repetition of the digits is not allowed?

Accepted Answer

The number of ways to arrange 5 distinct digits taken 3 at a time without repetition is given by $^5P3 = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 	imes 4 	imes 3 = 60$. Therefore, option B is correct.

Permutations and Combinations

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