Question:

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

Updated On: Oct 21, 2023
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Solution and Explanation

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=9P3^9P_3
=9!(93)!=\frac{9!}{(9-3)!}

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

=9×8×7×66!=\frac{9\times8\times7\times6}{6!}

=9×8×7=504=9\times8\times7=504

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Concepts Used:

Permutations and Combinations

Permutation:

Permutation is the method or the act of arranging members of a set into an order or a sequence. 

  • In the process of rearranging the numbers, subsets of sets are created to determine all possible arrangement sequences of a single data point. 
  • A permutation is used in many events of daily life. It is used for a list of data where the data order matters.

Combination:

Combination is the method of forming subsets by selecting data from a larger set in a way that the selection order does not matter.

  • Combination refers to the combination of about n things taken k at a time without any repetition.
  • The combination is used for a group of data where the order of data does not matter.