To keep from getting bogged down with the n's and r's
Let's write the "Pascal's triangle formula" using
letters different from n and r:
sCt + sC(t+1) = (s+1)C(t+1) <-- Pascal's triangle formula
X = nCr, Y = nCr+1, and Z = nCr+2
X + Y = nCr + nC(r+1) = (n+1)C(r+1)
Y + Z = nC(r+1) + nC(r+2)
Let s = n and t = r+1 in Pascal's triangle formula:
Y+Z = nC(r+1) + nC(r+1+1) = nC(r+1) + nC(r+2)
Adding the two equations together:
X + 2Y + Z = (n+1)C(r+1) + (n+1)C(r+2),
Now let s = n+1, and t = r+1 in Pascal's triangle formula:
(n+1)C(r+1) + (n+1)C(r+1+1) = (n+1+1)C(r+1+1)
which simplifies to
(n+1)C(r+1) + (n+1)C(r+2) = (n+2)C(r+2)
That left side turned out to be the same as the right side of
X + 2Y + Z above. So
X + 2Y + Z = (n+1)C(r+1) + (n+1)C(r+2) = (n+2)C(r+2)
or
X + 2Y + Z = (n+2)C(r+2)
To get X + Y + Z we need to subtract Y = nC(r+1) from
both sides:
X + 2Y + Z = (n+2)C(r+2)
Y = nC(r+1)
------------------------
X + Y + Z = (n+2)C(r+2) - nC(r+1)
Edwin