Question 940228
<pre>
nCr + nC(r+1) ?=? (n+1)C(r+1)
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(1)   nCr = {{{n!/(r!(n-r)!)}}}

(2)   nC(r+1) = {{{n!/((r+1)!(n^""-(r+1))!)}}} = {{{n!/((r+1)!(n-r-1)!)}}}

(3)   (n+1)C(r+1) = {{{(n+1)!/((r+1)!((n+1)^""-(r+1))!)}}} = {{{(n+1)!/((r+1)!(n+1-r-1)!)}}} = {{{(n+1)!/((r+1)!(n-r)!)}}}

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We want to prove that expression(1) + expression(2) = expression(3)

nCr + nC(r+1) = 

{{{n!/(r!(n-r)!)}}}{{{""+""}}}{{{n!/((r+1)!(n-r-1)!)}}}

Substitute (n-r)(n-r-1)! for (n-r)! and (r+1)r! for (r+1)!  

{{{n!/(r!(n-r)(n-r-1)!)}}}{{{""+""}}}{{{n!/((r+1)r!(n-r-1)!)}}} 

LCD = (r+1)r!(n-r)(n-r-1)!

{{{(n!(red(r+1)))/( (red(r+1))r!(n-r)(n-r-1)!)}}}{{{""+""}}}{{{n!(red(n-r))/((r+1)r!(red(n-r))(n-r-1)!)}}} 

Replace (r+1)r! by (r+1)! and replace (n-r)(n-r-1) by (n-r)!

{{{(n!(r+1))/( (r+1)!(n-r)!)}}}{{{""+""}}}{{{n!(n-r)/((r+1)!(n-r)!)}}} 

{{{(n!(r+1)+n!(n-r))/((r+1)!(n-r)!)}}} 

{{{(n!r+n!+n!n-n!r)/((r+1)!(n-r)!)}}}

{{{(cross(n!r)+n!+n!n-cross(n!r))/((r+1)!(n-r)!)}}}

{{{(n!+n!n)/((r+1)!(n-r)!)}}}

{{{(n!(1+n))/((r+1)!(n-r)!)}}}

{{{((n+1)n!)/((r+1)!(n-r)!)}}}

{{{(n+1)!/((r+1)!(n-r)!)}}}

(n+1)C(r+1)

Edwin</pre>