a,b,c,d are positive integers.find the no of possible solutions to the equation a+b+c+d=10

• Its the direct application of Stars & bars Rule.
  Now putting a=a-1,b=b-1,c=c-1,d=d-1
  The above equation reduce to a+b+c+d=10-4=6
  a+b+c+d=6
  Now there are 6 stars and 4-1(a,b,c,d) bars
  So the combination
  n+k-1 C k-1 = 6+(4-1) C 4-1
  = 9 C 3
  = 84
  
• 11 years agoHelpfull: Yes(42) No(6)
• 13C3 Formula : (n+r-1)C(r-1)
• 11 years agoHelpfull: Yes(13) No(5)
• IF given integers are whole numbers ans is 13c3
  if natural numbers ans is 9c3
• 11 years agoHelpfull: Yes(13) No(1)
• using following formula "(n+r-1)c(r-1)" to find out the result
  here r=4(a,b,c,d) and n=10......so
  subsuit (10+4-1)c(4-1)
  13c3=286
  so answer is 286
• 11 years agoHelpfull: Yes(7) No(6)
• 7,1,1,1 = 4 cases
  6,2,1,1 = 12 cases
  5,3,1,1 =12 cases
  5,2,2,1 = 12 cases
  4,3,2,1 = 24 cases
  4,4,1,1 = 6 cases
  4,2,2,2= 4 cases
  74 ans
  
• 11 years agoHelpfull: Yes(6) No(9)
• 9C3.......
  
• 11 years agoHelpfull: Yes(5) No(2)
• 9c3
  a,b,c,d are 4(a) terms...
  n we have to make der some 10(n)...
  so
  (n-1)c(a-1) will give you answer
  
• 11 years agoHelpfull: Yes(2) No(2)
• a+b+c+d=10 where a,b,c,d can even take whole no values.so to ease at taking natural no values we do:(a+1)+(b+1)+(c+1)+(d+1)=10+4
  W+X+Y+Z=14(is the same as previous,but here X,Y,Z and W take only natural no solutions)
  let us now take 14 as 14 empty seats and since the values of X,W,Y,z changes every instant,we can take operator(+) as the no of persons ,but since the operators to be put in between operands,we get 13 vacant places and these 3 are put in 13c3 ways in them
  eg: _ _ _ _ _ _ _ _ _ _ _ _ _ (spaces) + + + (operators) can be put in between in 13c3
• 11 years agoHelpfull: Yes(1) No(3)
• you can easily solve this question->
  since it is given that a, b,c,d are positive integers so we apply the formula n-1Cr-1 instead of n+r-1Cr-1 bcoz this formula is for non negative integers (including 0) hence answer is 9C3
• 11 years agoHelpfull: Yes(1) No(0)
• no sol
  bcoz there r 4 variabls...?
• 11 years agoHelpfull: Yes(1) No(1)
• 10+4-1C(4-1)=7C3=35
• 11 years agoHelpfull: Yes(0) No(6)
• If (a+b)=x ; (c+d)=y ; then the above equation can be represented as
  x+y=10 -----> This gives us Infinite solutions.
  Now If we consider a,b,c,d and equation a+b+c+d=10
  by considering the possible cases, this gives us 74 Positive solutions and
  78 Non-Negative Solutions.
• 11 years agoHelpfull: Yes(0) No(3)
• sateesh pls explain whyu hv used that formula
• 11 years agoHelpfull: Yes(0) No(0)
• 24 possible solutions.
  because 1+2+3+4=10 is the only possible situation as a,b,c,d are different positive integers.
  1,2,3,4 can be arranged in 4! ways.hence 24 solutions.
• 11 years agoHelpfull: Yes(0) No(0)
• plz...friends explain again....
• 11 years agoHelpfull: Yes(0) No(0)
• 24 solutions are there as
  1+2+3+4,1+3+2+4,1+4+2+3,1+3+4+2,1+2+4+3, and 1+4+3+2
  for 1 integer we have 6 solutions so for 4 integers it is 6*4 i.e, 24 solutions.
• 11 years agoHelpfull: Yes(0) No(0)
• formula
  "(n+r-1)c(r-1)"
  to find result
  here r=4(a,b,c,d) and n=10......so
  substitute (10+4-1)c(4-1)
  13c3=286
  
  286 is the answer
  
• 11 years agoHelpfull: Yes(0) No(1)
• 1+2+3+4 = 10
  So, total 4! possible no.of solutions available.
• 10 years agoHelpfull: Yes(0) No(0)