#1**-1 **

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.

atlas9 May 30, 2020

#3**0 **

You are adding two combinations: _{m}C_{k-1} + _{m}C_{k}

_{1) m}C_{k} = m! / [ k! · (m - k)! ]

_{2) m}C_{k-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 _{8}C_{3} + _{8}C_{2}

geno3141 Jun 2, 2020