how many 4-digits numbers are there such that the thousands digit is equal to the sum of other three digits?

• ans is 219
  when thousand digit is 1, 1100,1010,1001 are possible numbers=3
  when thousand digit is 2 possible cases are 6
  when thousand digit is 3 possible numbers are 10
  for 4=15
  for 5=21
  for 6=28
  for 7=36
  for 8=45
  for 9=55
  so total=3+6+10+15+21+28+36+45+55=219
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• Does anyone know the short cut method
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