# SOLUTION: How many 5 digit number exist between 10,500 and 11,000 if no digit may be repeated? I answered 120 - 5x4x3x2x1 = 120 on a quiz and it was counted incorrect. The teacher said t

Algebra ->  Algebra  -> Permutations -> SOLUTION: How many 5 digit number exist between 10,500 and 11,000 if no digit may be repeated? I answered 120 - 5x4x3x2x1 = 120 on a quiz and it was counted incorrect. The teacher said t      Log On

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 Click here to see ALL problems on Permutations Question 98520: How many 5 digit number exist between 10,500 and 11,000 if no digit may be repeated? I answered 120 - 5x4x3x2x1 = 120 on a quiz and it was counted incorrect. The teacher said that the answer was 210 but did not explain what I did wrong. Help! I'm trying to study for the test now and I can't figure it out.Answer by jim_thompson5910(28593)   (Show Source): You can put this solution on YOUR website!Let's look at the first digit: Since the number goes from 10,500 to 11,000, there is only one possibility for the first number: the only number it can be is "1" -------------------------------------------------------- Let's look at the second digit: Since the number goes from 10,500 to 11,000, there are two possibilities for the second number. The two possibilities are: 0 and 1 However since 1 is already taken, our only possibility for the second digit is 0 -------------------------------------------------------- Let's look at the third digit: Since the number goes from 10,500 to 11,000, we can have the numbers 5,6,7,8,9 (not zero since it's taken) for the third digit So we have 5 possibilities for the third digit -------------------------------------------------------- Let's look at the fourth digit: Now lets use any remaining digits to form the fourth possible digit. The remaining digits are: 2,3,4,5,6,7,8,9 Now lets say our third number is 5, so our possible numbers for the fourth digit is 2,3,4,6,7,8,9 (notice I've excluded 5) So we'll have 7 possibilities -------------------------------------------------------- Let's look at the fifth digit: Now lets use any remaining digits to form the fifth possible digit. The remaining digits are: 2,3,4,6,7,8,9 Now lets say our fourth number is 6, so our possible numbers for the fifth digit is 2,3,4,7,8,9 (notice I've excluded 6) So we'll have 6 possibilities Now multiply all of these combinations to get: 1*1*5*7*6=210 ------------------------------------------------------------------------------- You can also look at it like this: Since we're using a permutation, we can use nPr. So we have 7 possible numbers and we're choosing 3 digits. So we'll get Start with the given nPr formula Plug in n=7 and r=3 Subtract Expand Cancel like terms Simplify Now multiply 7*6*5 to get 210 Notice how this is identical to the previous explanation (the numbers are just in different order)