what are the number of solutions for x+y+z=12, if x,y & z can take values from 1 to 7 ?

• Ans=37
  if x=1, (y,z) can be (4,7)(5,6)(6,5)(7,4)
  if x=2, (y,z) can be (3,7)(4,6)(5,5)(6,4)(7,3)
  if x=3, (y,z) can be (2,7)(3,6)(4,5)(5,4)(6,3)(7,2)
  if x=4, (y,z) can be (1,7)(2,6)(3,5)(4,4)(5,3)(6,2)(7,1)
  if x=5, (y,z) can be (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)
  if x=6, (y,z) can be (1,5)(2,4)(3,3)(4,2)(5,1)
  if x=7, (y,z) can be (1,4)(2,3)(3,2)(4,1)
  Total Combination = 4+5+6+7+6+5+4=37
• 10 years agoHelpfull: Yes(68) No(2)
• Sorry I was wrong.Its 37 ways..
  Possiblities are
  1 4 7
  1 5 6
  2 3 7
  2 4 6
  3 4 5
  2 5 5
  3 3 6
  4 4 4
  
  These 7 cases will add up to 12.
  
  Btw take the first case 1 4 7
  x=1 y=4 z=7
  x=1 y=7 z=4
  x=4 y=1 z=7
  x=4 y=7 z=1
  x=7 y=4 z=1
  x=7 y=1 z=4
  Thus for the first case, 6 permutations are possible..
  
  for the first 5 cases , 6*5 =30 permutations are possible.
  
  But for 3 3 6 and 2 5 5 repetition is there. (3!/2!)*2 = 6
  
  And for 4 4 4, 1 permutation.
  
  So
  30+6+1=37
• 10 years agoHelpfull: Yes(10) No(3)
• The correct answer is 37.
• 10 years agoHelpfull: Yes(4) No(1)
• I think 42 ways.. I couldn't solve in any easy method.
  Possiblities are
  1 4 7
  1 5 6
  2 3 7
  2 4 6
  2 5 5
  3 3 6
  3 4 5
  
  These 7 cases will add up to 12.
  
  Btw take the first case 1 4 7
  x=1 y=4 z=7
  x=1 y=7 z=4
  x=4 y=1 z=7
  x=4 y=7 z=1
  x=7 y=4 z=1
  x=7 y=1 z=4
  Thus for the first case, 6 permutations are possible..
  we have 7 cases,so 6*7=42
• 10 years agoHelpfull: Yes(3) No(15)
• I made a mistake...37 is correct answer
• 10 years agoHelpfull: Yes(1) No(2)
• we can write as 7+4+1/7+3+2 smilarly we can write n get 9 solutn
• 10 years agoHelpfull: Yes(0) No(7)
• 12+3-1 C 12=91.
  
• 10 years agoHelpfull: Yes(0) No(10)
• 43 solutions are possible
• 10 years agoHelpfull: Yes(0) No(12)
• (1,4,7)---->can be arranged in 6 ways to give a solution
  (1,5,6)---->can be arranged in 6 ways to give a solution
  (2,3,7)---->can be arranged in 6 ways to give a solution
  (2,4,6)---->can be arranged in 6 ways to give a solution
  (2,5,5)---->can be arranged in 3 ways to give a solution
  (3,3,6)---->can be arranged in 3 ways to give a solution
  (3,4,5)---->can be arranged in 6 ways to give a solution
  (4,4,4)---->can be arranged in 1 way to give a solution
  (5,5,2)---->can be arranged in 3 ways to give a solution
  
  Total no of ways= 40
• 10 years agoHelpfull: Yes(0) No(14)
• sorry for previous answer it will be 37
• 10 years agoHelpfull: Yes(0) No(3)
• @KARTHIKA why didnt u take 5 5 2???
• 10 years agoHelpfull: Yes(0) No(1)
• @Rudra:
  x=2 y=5 z=5
  x=5 y=5 z=2
  x=5 y=2 z=5
  these are the three ways...
  So I just tuk 2 5 5 and did permutation. i.e 3!/2!=3ways as mentioned in detail above (Dividing by 2!,because "5" is repeated two times)
  Rudra hope yu understud
  
• 10 years agoHelpfull: Yes(0) No(0)