How many ways can you write 410 as the sum of 4 different 3-digit numbers, assumingorder does not matter?
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 \(H^4_{10} =\displaystyle \binom{10 + 4-1}{10} = \binom{13}{10}=286\)