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Worked Examples · Example 2

Q.How many 3-digit numbers can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated?

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✓ Free question

We must fill three positions — hundreds, tens, units — using the 5 given digits, with no digit repeated.

Hundreds place: Any of the 5 digits can go here ⇒5\Rightarrow 5 ways.

Tens place: One digit has already been used, so 44 digits remain ⇒4\Rightarrow 4 ways.

Units place: Two digits have already been used, so 33 digits remain ⇒3\Rightarrow 3 ways.

By the multiplication principle, the total number of 3-digit numbers is

5×4×3=605 \times 4 \times 3 = 60

Verification using the permutation formula: This is exactly 5P3^{5}P_{3} (arranging 3 out of 5 distinct digits, order mattering since digit position matters):

5P3=5!(5−3)!=5!2!=1202=60^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{120}{2} = 60

Both methods agree: 60.

✓Final answer

60 three-digit numbers.

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