+261 could you show what it looked like exactly?
if it's what you wrote right now then just combine like terms:
(mk-1)+(mk)
=mk-1+mk
=2mk-1
you didn't need the parentheses at the beginning because of the associative property i think.
+23246 You are adding two combinations: mCk-1 + mCk
1) mCk = m! / [ k! · (m - k)! ]
2) mCk-1 = m! / [ (k - 1)! · ( m - (k - 1) )! ] = m! / [ (k - 1)! · ( m - k +1 )! ]
To add them, they must have common denominators.
Formula 1) needs the term (m - k + 1) in its denominator because (m - k + 1)! = (m - k + 1) ·(m - k)!
Formula 2) needs the term k in its denominator because k! = k ·(k - 1)!
So: 1) m! / [ k! · (m - k)! ] · [ (m - k + 1) / (m - k + 1) ] = [ m!·(m - k + 1) ] / [ k! · (m - k + 1 )! ]
2) m! / [ (k - 1)! · ( m - k +1 )! ] · [ k / k ] = [ m! · k ] / [ k! · (m - k + 1)! ]
Adding the numerators of these two parts: [ m!·(m - k + 1) ] + [ m! · k ] = m! · [ (m - k + 1) + k ]
= m! · (m + 1)
The full fraction: [ m! · (m + 1) ] / [ k! · (m - k + 1)! ] = (m + 1)! / [ k! · (m - k + 1)! ]
Probably, to make sense of this, try some particular problems, such as 8C3 + 8C2
