We can see here that there are pairs of binomial coefficients involved in the expression so we need to multiply two binomial expansions, the simplest one being \((1+ x)^{n}\).

\((1+ x)^{n}\) = \({n \choose 0} + {n \choose 1} x + {n \choose 2} x ^ {2} + ..... + {n \choose n-1} x ^ {n-1} + {n \choose n} x ^ {n}\)

\((x + 1)^{n}\) = \({n \choose 0} x ^ {n} + {n \choose 1} x ^ {n-1}+ {n \choose 2} x ^ {n-2} + ..... + {n \choose n-1} x + {n \choose n} \)

Multiplying both we get,

\((1+ x)^{2n}\) = \({n \choose 0}.{n \choose 1} x ^ {n-1} + {n \choose 1}. {n \choose 2} x ^ {n-1} + .... + {n \choose n-1}.{n \choose n}x ^ {n-1} + ..... \) many other terms.

Every term in the above expression has \( x ^ {n-1}\).
So we can say that the sum of the coefficients in the above expression is equal to the coefficient of \( x ^ {n-1}\) on the left hand side i.e. in \((1+ x)^{2n}\) which is of course \({2n \choose n-1}\).

You can also generalize your result after multiplying the two equation by equating the coefficient of \(x^{n-r}\) from both sides to get \[{n \choose 0} . {n \choose r} + {n \choose 1} . {n \choose r+1}+ \ldots + {n \choose n-r }.{ n \choose n} = {2n \choose n-r} \]

How would one prove Vandermonde's identity (http://en.wikipedia.org/wiki/Vandermonde's_identity) using this method? It looks like it would be derivable in the same way.

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TopNewestWe can see here that there are pairs of binomial coefficients involved in the expression so we need to multiply two binomial expansions, the simplest one being \((1+ x)^{n}\).

\((1+ x)^{n}\) = \({n \choose 0} + {n \choose 1} x + {n \choose 2} x ^ {2} + ..... + {n \choose n-1} x ^ {n-1} + {n \choose n} x ^ {n}\)

\((x + 1)^{n}\) = \({n \choose 0} x ^ {n} + {n \choose 1} x ^ {n-1}+ {n \choose 2} x ^ {n-2} + ..... + {n \choose n-1} x + {n \choose n} \)

Multiplying both we get,

\((1+ x)^{2n}\) = \({n \choose 0}.{n \choose 1} x ^ {n-1} + {n \choose 1}. {n \choose 2} x ^ {n-1} + .... + {n \choose n-1}.{n \choose n}x ^ {n-1} + ..... \) many other terms.

Every term in the above expression has \( x ^ {n-1}\). So we can say that the sum of the coefficients in the above expression is equal to the coefficient of \( x ^ {n-1}\) on the left hand side i.e. in \((1+ x)^{2n}\) which is of course \({2n \choose n-1}\).

Therefore,

\({n \choose 0}.{n \choose 1} + {n \choose 1}. {n \choose 2} + ...... + {n \choose n-1}.{n \choose n} = {2n \choose n-1}\)

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You can also generalize your result after multiplying the two equation by equating the coefficient of \(x^{n-r}\) from both sides to get \[{n \choose 0} . {n \choose r} + {n \choose 1} . {n \choose r+1}+ \ldots + {n \choose n-r }.{ n \choose n} = {2n \choose n-r} \]

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How would one prove Vandermonde's identity (http://en.wikipedia.org/wiki/Vandermonde's_identity) using this method? It looks like it would be derivable in the same way.

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wow ... if you don't mind me asking, roughly how long does it take you to work these kind of things out? o.o

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Well, i had this question before

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How do I find the sum of this series :-

(nC2)(nC1)(nC0)+(nC3)(nC2)(nC1)+(nC4)(nC3)(nC2)+.....................+(nCn)(nC(n-1))(nC(n-2))

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