Mathematics Homework Solutions
Problem
#8752

One equality

Hello!

The attached file is the answer an OTA gave me to a problem I was having. There is one line in the reasoning I do not understand, and I have highlighted it in red.

I would be very grateful if someone could explain it to me.

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s5.doc
Problem:

I’m trying to prove the following:



using the following identity:



I have tried various methods, but I am baffeled!

Incidentally, I would be very grateful if you could also show me how to
prove the second formula.

Solution:

We can solve following identity without using the second one.

First we have to expand the left-hand side term by using the formula

nCr = (n)(/[(r) ( (n-r) (]

Applying in left-hand side we will get

LHS = n-1Ck-m-1*n-1Cm-1 + n-1Ck-m-2*n-1Cm +n-1Ck-m-3*n-1Cm+1 +…….

= (n-1)(/(k-m-1)((n-k+m)( * (n-1)(/(m-1)((n-m)( +
(n-1)(/(n-k+m+1)((k-m-2)( *

(n-1)(/m((n-1-m)( +(n-1)(/(k-m-3)(( n-k+m+2)( *(n-1)(/(m+1)( (n-m-2)(
+……

= (n-1)(*(n-1)( { 1/[(k-m-1)((n-k+m)( (m-1)((n-m)( ] + 1/[
(n-k+m+1)((k-m-2)( m(

(n-1-m)( ] + 1/[(k-m-3)(( n-k+m+2)( (m+1)( (n-m-2)(] +……..

taking (m+n-k)( and (k-1)( common from denominator

finally we will get

1/[(k-1)( (m+n-k)(]{(n-1)( *(n-1+1)(n-1+2)(n-1+3)……..(n-1+m)} How
does this follow from the previous line?

= 1/[(k-1)( (m+n-k)(]{(m+n-1)()}

=(m+n-1)(/[(k-1)( (m+n-k)(]

= m+n-1Ck-1
Solution
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