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doubt of numbers
by deepak kushwaha - Saturday, 21 July 2012, 06:46 PM
  How many ordered triplets (a, b, c) of positive odd integers satisfy a + b + c = 23?
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Re: doubt of numbers
by rakshit surana - Saturday, 21 July 2012, 08:25 PM
  let a = 2k - 1
b = 2k` - 1
c = 2k`` - 1
where k, k` , k`` >=1
substituting
2k + 2k` + 2k`` - 3 = 23
2( k + k` + k`` ) = 26
k + k` + k`` = 13
number of poitive solutions is given by C(n-1, r-1)
C(12, 2) = 66 ans
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Re: doubt of numbers
by harendra chaudhari - Monday, 23 July 2012, 10:33 AM
  a+b+c=23
let a=2p+1, b=2q+1, 2r+1
2p+2q+2r+3=23
2p+2q+2r=20
p+q+r=10
positive integral solutions=C(10,3)
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Re: doubt of numbers
by TG Team - Monday, 23 July 2012, 12:33 PM
 

It is C(12, 2) = 66.

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Re: doubt of numbers
by rakshit surana - Monday, 23 July 2012, 12:43 PM
  hi harendra, i think u did a lil mistake
since 2p +1 u have taken p will take all values including 0
p+q+r = 10
no of non negative solutions is C(n+r-1, r-1)
so C(10+3-1, 3-1)
C(12,2) which is equivalent to my previous solution
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