Summing Combinatorics

How can I find the value of the following series ::--

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

#Combinatorics #HelpMe! #MathProblem #Math

Note by Sanjay Banerji
7 years, 6 months ago

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12 votes

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Comments

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}.

(1+x)n(1+ x)^{n} = (n0)+(n1)x+(n2)x2+.....+(nn−1)xn−1+(nn)xn{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(x + 1)^{n} = (n0)xn+(n1)xn−1+(n2)xn−2+.....+(nn−1)x+(nn){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(1+ x)^{2n} = (n0).(n1)xn−1+(n1).(n2)xn−1+....+(nn−1).(nn)xn−1+.....{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 xn−1 x ^ {n-1}. So we can say that the sum of the coefficients in the above expression is equal to the coefficient of xn−1 x ^ {n-1} on the left hand side i.e. in (1+x)2n(1+ x)^{2n} which is of course (2nn−1){2n \choose n-1}.

Therefore,

(n0).(n1)+(n1).(n2)+......+(nn−1).(nn)=(2nn−1){n \choose 0}.{n \choose 1} + {n \choose 1}. {n \choose 2} + ...... + {n \choose n-1}.{n \choose n} = {2n \choose n-1}

Shaan Vaidya - 7 years, 6 months ago

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You can also generalize your result after multiplying the two equation by equating the coefficient of xn−rx^{n-r} from both sides to get (n0).(nr)+(n1).(nr+1)+…+(nn−r).(nn)=(2nn−r){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}

Snehdeep Arora - 7 years, 6 months ago

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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.

Michael Tang - 7 years, 6 months ago

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

Jord W - 7 years, 6 months ago

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Well, i had this question before

Shaan Vaidya - 7 years, 6 months ago

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@Shaan Vaidya – oh right .. well cool anyway

Jord W - 7 years, 6 months ago

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))

Sanjay Banerji - 7 years, 6 months ago
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