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How many 3 - digit positive integers can be formed by using 1, 2, 3, 4, 5 without the repetition of digits?

 Jan 3, 2020
 #1
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Use constructive counting.

 Jan 3, 2020
 #2
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Use the permutations formula:

 

5 nPr 3 = 60

 

5! / (5 - 3)! =120 / 2 = 60 such numbers

 Jan 3, 2020
 #3
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use constructive counting and just have patience and try, that's how i do it.

 Jan 3, 2020
 #4
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5  choices for first digit leaves

4 choices for the second digit leaves

3 choices for the third digit....

  5 x 4 x 3 = 60 possibilities

 Jan 3, 2020

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