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Question
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
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Solution
3-digit numbers have to be formed using the digits 1 to 9.
Here, the order of the digits matters.
Therefore, there will be as many 3-digit numbers as there are permutations of 9 different digits taken 3 at a time.
Therefore, required number of 3-digit numbers
= 9P3 = `(9!)/((9 - 3)!) = (9!)/(6!)`
= `(9 xx 8 xx 7 xx 6!)/(6!)`
= 9 x 8 x 7 = 504
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