How many 3-digit numbers can be formed using 5, 6, 7 and 8 if the digits cannot be repeated?

Question

How many 3-digit numbers can be formed using 5, 6, 7 and 8 if the digits cannot be repeated?

Answer 1

first digit: 4 choices from 5,6,7 or 8

second digit: 3 choices remain because of non-repetition stricture

third digit: 2 choices remain

 

Product of the above is 4! = 24

Answer 2

Let’s imagine this as an experiment:

  There are three digits.

  The first one may be any of the 4 possibilities (it does not matter that they are numbers).

  Then, the second one may be any of the remaining 3 possibilities.

  Then, the third one may be any of the remaining 2 possibilities.

 

We consider 567 and 576 to be different 3-digit numbers, so they are both counted.

 

This gives 4*3*2 = 24 possible 3-digit numbers.  You can list them pretty easily:

     FOR A= 5 TO 8

        FOR B=5 TO 8

          IF (B<>A) THEN

            FOR C= 5 to 8

              IF ((C<>A) AND (C<>B)) THEN OUTPUT (A,B,C)

            NEXT C

         NEXT B

      NEXT A