33 coins are there which are to be distributed among 3 sons A, B, C. No son gets more than 15, since C is youngest, he gets more than A or B & they are in arithematic progression. How many distributions are possible

• A B C
  7 11 15
  8 11 14
  9 11 13
  10 11 12
  hence 4 distribution.
• 11 years agoHelpfull: Yes(36) No(9)
• C A B
  15 11 7
  15 7 11
  14 11 8
  14 8 11
  13 9 11
  13 11 9
  12 11 10
  12 10 11
  So possibilities are 8
  bcz we dont know whether a is younger or b is younger
  
• 11 years agoHelpfull: Yes(26) No(12)
• The answer is 8
  
  for C 15, then there are 2 chances like C=15, B=11, A=7 if B is younger than A, or C=15, A=11, B=7, if A is younger than B, similarly for all others.
  This is because we dont know who is younger among A & B
• 11 years agoHelpfull: Yes(20) No(4)
• let the share of youngest be n
  and the difference of AP be r
  the shares of A B and C are
  n-2r, n-r, n
  sum= (n-2r) + (n-r) + (n) =3*(n-r)=33
  r=n-11 must atleast 1.
  n can be more than 15.
  n=12,13,14,15
• 11 years agoHelpfull: Yes(10) No(0)
• let a,b,c, be in A.P so b=a+c/2 also c>a,b and a,b,ca so possibilities are
  A B C
  7 11 15
  8 11 14
  9 11 13
  10 11 12
• 11 years agoHelpfull: Yes(5) No(0)
• hence it is in a.p so the minimum diff ll be 1.if so then
  A B C
  10 11 12
  now c may varies between 12-15 i.e. 12,13,14,15 so that the distribution becomes
  A B C
  9 11 13 diff is 2
  8 11 14 diff is 3
  7 11 15 diff is 4
  hence 4 distribution.
• 10 years agoHelpfull: Yes(3) No(1)
• I think the answer is 8 as follows
  
  15 for C & 7, 11 for A & B in 2 ways
  14 for C & 8, 11 for A & B in 2 ways
  similarly for 13 for C ... & 12 for C ....
• 11 years agoHelpfull: Yes(1) No(3)
• By AP,
  
  a,a+d,a+2d=33
  wkt,a=15
  so,15,15+d,15+2d=33
  45+3d=33
  3d=12
  d=4
  so 4 is the answer...
• 10 years agoHelpfull: Yes(1) No(1)
• 8 ways.. 2!*4 ways as either of A and B can get more coins..
• 10 years agoHelpfull: Yes(1) No(1)
• 11 distributions are possible
• 10 years agoHelpfull: Yes(1) No(0)
• to those who thinking this question is a hard one---
  among A,B and C ....C will always get more than that of other two,and it is given that the max value of C can be 15,therefore,keeping in mind that c is always greater ,,thus when the value of C will reach to 11(decrementing by one unit value),,the solution will no longer be possible as the value of A and B becomes greater..hence 8 is the possible solutions that we can derive.:)
• 9 years agoHelpfull: Yes(1) No(3)
• only 4 combination is possible bcoz they are in AP.
  A B C
  7 11 15
  8 11 14
  9 11 13
  10 11 12
• 8 years agoHelpfull: Yes(1) No(0)
• C:13
  B:11
  A:9
• 11 years agoHelpfull: Yes(0) No(3)
• 26 distributions
• 11 years agoHelpfull: Yes(0) No(3)
• 9,11,13
  coz 9,9+2,9+4
• 11 years agoHelpfull: Yes(0) No(1)
• x+x+d+x+2d=33
  x+d=11
  so,x=10, x+d=11, x+2d=12
• 9 years agoHelpfull: Yes(0) No(0)
• suppose a-d ,a and a+d are coins
  
  according to question (a-d)+a+(a+d)=33 => a=11
  now d may be 1,2,3,4 in order to a+d
• 9 years agoHelpfull: Yes(0) No(0)
• 8 distributions..according to given condition ..because it is not mention the younger in A &B so we take both the possibilities...
• 7 years agoHelpfull: Yes(0) No(0)