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

Asked by Pragya Singh | 1 year ago |  89

##### Solution :-

Here, what matters is the order of digits. Thus, there are permutations of 9 different digits taken 3 at a time.

Therefore, the number of 3-digit numbers =

$$^9P_3=\dfrac{9!}{(9-3)!}=\dfrac{9!}{6!}$$

$$\dfrac{9\times 8\times 7\times 6!}{6!}$$

= 9×8×7=504

Answered by Abhisek | 1 year ago

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