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How many ways can you write 410 as the sum of 4 different 3-digit numbers, assumingorder does not matter?

 Jan 28, 2021
 #1
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Each number is a 3-digit number.

 

This means a + b + c + d = 410 where a,b,c,d >= 100.

 

Let a' = a - 100, b' = b - 100, etc.

 

Then a' + b' + c' + d' = 10, where a,b,c,d >= 0.

 

Then the required number of ways is \(H^4_{10} =\displaystyle \binom{10 + 4-1}{10} = \binom{13}{10}=286\)

 Jan 28, 2021

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