IF x, y and z can only take values 1,2,3,4,5,6 and 7. Then find the number of possible solutions of the equation x +y +z =12.

• 37 solutions
  
  If repetition of numbers is allowed, then possible solutions with x+y+z=12 are
  (1,4,7) (1,5,6) (1,6,5) (1,7,4)
  (2,3,5) (2,4,6) (2,5,5) (2,6,4) (2,7,3)
  (3,2,7) (3,3,6) (3,4,5) (3,5,4) (3,6,3) (3,7,2)
  (4,1,7) (4,2,6) (4,3,5) (4,4,4) (4,5,3) (4,6,2) (4,7,1)
  (5,1,6) (5,2,5) (5,3,4) (5,4,3) (5,5,2) (5,6,1)
  (6,1,5) (6,2,4) (6,3,3) (6,4,2) (6,5,1)
  (7,1,4) (7,2,3) (7,3,2) (7,4,1)
  =Total 37
• 9 years agoHelpfull: Yes(0) No(0)
• In second row, pl. read first solution as (2,3,7) instead of (2,3,5).
• 9 years agoHelpfull: Yes(0) No(0)