×

# Summing Combinatorics

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

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

Note by Sanjay Banerji
3 years, 8 months ago

Sort by:

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 $${2n \choose n-1}$$.

Therefore,

$${n \choose 0}.{n \choose 1} + {n \choose 1}. {n \choose 2} + ...... + {n \choose n-1}.{n \choose n} = {2n \choose n-1}$$ · 3 years, 8 months ago

You can also generalize your result after multiplying the two equation by equating the coefficient of $$x^{n-r}$$ from both sides to get ${n \choose 0} . {n \choose r} + {n \choose 1} . {n \choose r+1}+ \ldots + {n \choose n-r }.{ n \choose n} = {2n \choose n-r}$ · 3 years, 8 months ago

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. · 3 years, 8 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 · 3 years, 8 months ago

Well, i had this question before · 3 years, 8 months ago

oh right .. well cool anyway · 3 years, 8 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)) · 3 years, 8 months ago