Subsets having three natural numbers are being made from the set of natural numbers, 1 to 63 (including both). Prove that number of such subsets for which (a+b+c)<95 is less than the number of subsets having (a+b+c)>95. a,b and c are notations used for three natural numbers in the subsets. The subsets have distinct numbers.Krishna

Heres a quick stab at it :-)let a,b,c be disticnt members of {1,2,...,62,63}a+b+c<n => -(a+b+c)>-n => (64 - a) + (64 - b) + (64 - c) >192 -n => (x+y+z)>192-n where x,y,z, are distinct members of (1,2,...,62,63} (clearly (a,b,c) --> (x,y,z) is a one to one function).So, clearly, the number of subsets for which (a+b+c)<95 = the number of subsets for which (a+b+c) >192 - 95=97so #{a+b+c<95)=#{a+b+c>97}=#{a+b+c>95}-#{a+b+c=96}-#{(a+b+c=97}<#{a+b+c>95}where the last inequality can be checked by seeing that e.g. (30+30+37)=97, so #{a+b+c=96}+#{(a+b+c=97} >0

Here is VBA-function that allows you to generalize this to n-tuplets of numbers <=M.Function ntuplets(M As Integer, n As Integer, p As Integer) As Integer'Counts the number of n-tuplets of natural numbers less than or equal to M whose sum is stricly less than p. Dim itemp As Integer Dim ntup As Variant Dim sum As Integer Dim antal As Long Dim changed As Boolean Dim i As Long Dim j As Long Dim l As Long ReDim ntup(1 To n) As Integer ReDim startvector(1 To n) As Integer antal = 1 For i = 1 To n ntup(i) = i antal = antal * (M - i + 1) / i Next i itemp = 0 i = 1 While i < antal + 1 sum = 0 For j = 1 To n sum = sum + ntup(j) Next j If sum < p Then itemp = itemp + 1 End If changed = False While Not changed And i < antal If ntup(n) < M Then ntup(n) = ntup(n) + 1 changed = True Else l = 2 While l < n + 1 And Not changed If ntup(n - l + 1) < M - (l - 1) Then ntup(n - l + 1) = ntup(n - l + 1) + 1 changed = True For j = 1 To l - 1 ntup(n - l + j + 1) = ntup(n - l + j) + 1 Next j Else l = l + 1 End If Wend End If Wend i = i + 1 Wend ntuplets = itempEnd FunctionRegards,Niclas

just by a symmetry argument and since (64/2)*3= 96, I would come to the same conclusion as Halliron that Quoteclearly, the number of subsets for which (a+b+c)<95 = the number of subsets for which (a+b+c) >192 - 95=97"and the remainder is left to the reader."

There are exactly 19,135 triplets for which (a+b+c)<95, 20,096 for which (a+b+c)>95 and 480 for which (a+b+c)=95.(And one can check that 19,135 + 20,096 + 480 = 39,711 = 63!/(60! * 3!) )