User talk:Melink14

Formula	Explanation
${\frac {n!}{k!(n-k)!}}+{\frac {n!}{(k-1)!(n-k+1)!}}$	Problem as stated
${\frac {n!}{k(k-1)!(n-k)!}}+{\frac {n!}{(k-1)!(n-k+1)(n-k)!}}$	Using rules of factorials, change denominators to make LCD finding easier
${\frac {n!(n-k+1)}{k(k-1)!(n-k+1)(n-k)!}}+{\frac {n!(k)}{k(k-1)!(n-k+1)(n-k)!}}$	Find least common denominator
${\frac {n!(n-k+1)+n!(k)}{k(k-1)!(n-k+1)(n-k)!}}$	Add the two fractions together
${\frac {n!(n-k+1+(k))}{k(k-1)!(n-k+1)(n-k)!}}$	Factor out the $n!$ in the numerator. This is where the + k comes from.
${\frac {(n+1)!}{k!(n+1-k)!}}$	Simplify.

As noted, this can also be written as ${\binom {n+1}{k}}$ .
--Melink14 (talk) 02:12, 26 April 2009 (UTC)[reply]

Now for the second question.

Formula	Explanation
${\binom {n}{k}}+{\binom {n}{k+1}}$	The stated problem.
${\frac {n!}{k!(n-k)!}}+{\frac {n!}{(k+1)!(n-k-1)!}}$	Write out as per the definition of combinations.
${\frac {n!}{k!(n-k-1)!(n-k)}}+{\frac {n!}{k!(k+1)(n-k-1)!}}$	Use factorial math to simplify denominators for LCD finding
${\frac {n!(k+1)}{k!(n-k-1)!(n-k)(k+1)}}+{\frac {n!(n-k)}{k!(n-k-1)!(n-k)(k+1)}}$	Find the LCD.
${\frac {n!(k+1)+n!(n-k)}{k!(n-k-1)!(n-k)(k+1)}}$	Add.
${\frac {n!(k+1+n-k)}{k!(n-k-1)!(n-k)(k+1)}}$	Factor out the $n!$ like last time.
${\frac {n!(n+1)}{k!(n-k-1)!(n-k)(k+1)}}$	Basic simplification of numerator
${\frac {(n+1)!}{(k+1)!(n-k)!}}$	Simplify again using rules of factorials
${\frac {(n+1)!}{(k+1)!(n-k)!}}={\frac {(n+1)!}{(k+1)!((n+1)-(k+1))!}}$	A simple identity to make the connection to the next part clear.
${\frac {(n+1)!}{(k+1)!((n+1)-(k+1))!}}={\binom {n+1}{k+1}}$	This is the final result the question was looking for.