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 ->  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


   

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(35256) About Me  (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"


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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

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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

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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

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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



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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

nPr=%28n%21%29%2F%28n-r%29%21 Start with the given nPr formula


7P3=%287%21%29%2F%287-3%29%21 Plug in n=7 and r=3


7P3=%287%21%29%2F%284%21%29 Subtract



7P3=%287%2A6%2A5%2A4%2A3%2A2%2A1%29%2F%284%2A3%2A2%2A1%29 Expand


7P3=%287%2A6%2A5%2Across%284%2A3%2A2%2A1%29%29%2F%28cross%284%2A3%2A2%2A1%29%29 Cancel like terms

7P3=7%2A6%2A5 Simplify

7P3=210 Now multiply 7*6*5 to get 210



Notice how this is identical to the previous explanation (the numbers are just in different order)