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Number of digits available = 9 (1, 2, 3, 4, 5, 6, 7, 8, 9) Number of digits to be used = 3 (No digit is to repeat) n = 9, r = 3 Number of permutations = Hence, the number of 3-digit numbers formed = 504. - Wired Faculty

Permutations And Combinations

Question

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How many 3-digit numbers can be formed by using digits 1 to 9 if no digit is repeated?

Solution

Number of digits available = 9
(1, 2, 3, 4, 5, 6, 7, 8, 9)
Number of digits to be used = 3
(No digit is to repeat)

$rightwards double arrow$ n = 9, r = 3
Number of permutations = $space space space straight P presuperscript 9 subscript 3 space equals space fraction numerator 9 factorial over denominator 6 factorial end fraction space equals space fraction numerator 9 cross times 8 cross times 7 cross times 6 factorial over denominator 6 factorial end fraction equals 504$
Hence, the number of 3-digit numbers formed = 504.

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Permutations And Combinations

How many 3-digit numbers can be formed by using digits 1 to 9 if no digit is repeated?

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