If a+b+c=21, what is the total number of non-negative integral solutions?

• Easy Way. Think of it is a problem of diving 21 items to three people A, B & C which may or may not receive an item.
  
  So simply use Combination here as (n-r+1) C (r-1) i.e.
  
  (21+3-1) C (3-1) => 23 C 2 == > 253 solutions.
  
• 8 years agoHelpfull: Yes(27) No(6)
• divide 21 items to 3 people A, B & C which may or may not receive an item.
  So simply use Combination here as (n+r-1) C (r-1) i.e.
  
  (21+3-1) C (3-1) => 23 C 2 == > 253 solutions.
• 8 years agoHelpfull: Yes(9) No(1)
• see this link, u get a clear idea abt it!!!
  http://www.careerbless.com/aptitude/qa/permutations_combinations_imp8.php
• 8 years agoHelpfull: Yes(7) No(0)
• 253 is the right ans wer but I didnt understand the solution.
• 8 years agoHelpfull: Yes(4) No(0)
• Ans= 28
  
  fix value of a from 3 to 9 such as:
  if a=3; then b+c=18; =) b=c=9 i.e. 1 way
  if a=4; then b+c=17; =) b=8/9; c=8/9 i.e. 2 ways
  
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  if a=9; then b+c=12;
  
  you will get total of 28 non-negative integral solutions
• 8 years agoHelpfull: Yes(1) No(4)
• 22C2
  =>231
• 8 years agoHelpfull: Yes(1) No(2)
• We can also solve this through partition method (this is method is used for finding the no. of solutions or distribution of identical things :
  a + b + c = 21
  
  the question said that we have separate this value among 3 (a, b, c)
  
  then,
  
  a + b + c =21!+2p! : p=partition ( we take 2p because we have to separate 21 in three(a , b , c)).
  
  a + b + c = 23! / 21! * 2! = 253 answer.
  
  (It is much simpler rather to remembering the formulae).
  
• 8 years agoHelpfull: Yes(0) No(0)
• Since, a+b+c=21
  Now for integral solution a>=1;b>=1;c>=1
  therefore, a=1+i1;b=1+i2;c=1+i3;
  again, i1+i2+i3=21-3=19=n;r=3
  using, (n+r-1)C(r-1)=253
  
• 8 years agoHelpfull: Yes(0) No(0)
• 22+21+20+........+2+1 = 253
• 8 years agoHelpfull: Yes(0) No(1)
• (n+r-1)C(r-1). Here, n=21 and r=3.
• 8 years agoHelpfull: Yes(0) No(0)