Re: discrete math with binomial theorem.


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"mina_world" <mina_world@xxxxxxxxxxx> wrote in message news:fahge8$q9i$1@xxxxxxxxxxxxxxxxxxx
Hello sir~

mC0.nCr + mC1.nC(r-1) + .... + mCr.nC0 = (m+n)Cr.

It's easy. Because,
(m+n)Cr is the coefficient of x^r in (1+x)^(m+n).
and
(1+x)^(m+n) = {(1+x)^m}{(1+x)^n}.
and
x^r = x^0.x^r = x^1.x^(r-1) = .... = x^r.x^0.

Since the coefficient of x^r in (1+x)^(m+n) and {(1+x)^m}{(1+x)^n},
mC0.nCr + mC1.nC(r-1) + .... + mCr.nC0 = (m+n)Cr.
-------------------------------------------------------------

mC0.nCr + mC1.nC(r+1) + .... + mCm.nC(r+m) = (m+n)C(m+r).

I don't know it well.
How can you explain it ?

Hint: mCi = mC(m-i).

By your hint,
mC0.nCr + mC1.nC(r+1) + .... + mCm.nC(r+m) = (m+n)C(m+r).
<=>
mC0.nC(n-r) + mC1.nC(n-r-1) + ....+ mCm.nC(n-r-m) = (m+n)C(n-r).

Maybe, n > r+m ==> n-r > m.

Sorry. n >= r+m ==> n-r >= m.

mC0.nCr + mC1.nC(r+1) + .... + mCm.nC(r+m)
= mCm.nCr + mC(m-1).nC(r+1) +...+ mC0.nC(r+m)
= (m+n)C(m+r)


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