Q

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

Q.1.    How many 3-digit numbers can be formed by using the digits 1 to 9

if no digit is repeated?

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3-digit numbers have to be formed by using the digits 1 to 9.

Here, the order of 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, the required number of 3-digit numbers $=^{9}P_3$

$=\frac{9!}{(9-3)!}$

$=\frac{9!}{6!}$

$=\frac{9\times 8\times 7\times 6!}{6!}$

$=9\times 8\times 7=504$

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