Question 98520
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=(n!)/(n-r)!}}} Start with the given nPr formula



{{{7P3=(7!)/(7-3)!}}} Plug in n=7 and r=3



{{{7P3=(7!)/(4!)}}} Subtract




{{{7P3=(7*6*5*4*3*2*1)/(4*3*2*1)}}} Expand



{{{7P3=(7*6*5*cross(4*3*2*1))/(cross(4*3*2*1))}}}  Cancel like terms


{{{7P3=7*6*5}}}  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)