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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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