Double summation

Probability Level 2

Find the value of - $\underset { 0\le i<j\le n }{ \sum { \sum { { C }_{ i }{ C }_{ j } } } }$

where $C_k = \dbinom n{k} = \dfrac{n!}{(n-k)! k!}$ .

{ 2 }^{ 2n-1 }-\frac { (2n)! }{ 2(n-1!)(n_1!) }

{ 2 }^{ 2n-1 }-\frac { (2n)! }{ (n!)(n!) }

{ 2 }^{ 2n }-\frac { (2n)! }{ (n!)(n!) }

{ 2 }^{ 2n-1 }-\frac { (2n)! }{ 2(n!)(n!) }

1 solution

Prajwal Krishna
Nov 6, 2016

[ $_{ }^{ n }{ C }_{ 0 }^{ }$ + $_{ }^{ n }{ C }_{ 1 }^{ }$ + $_{ }^{ n }{ C }_{ 2 }^{ }$ +.......+ $_{ }^{ n }{ C }_{ 0n}^{ }$ ] $\overset { 2 }{ }$ =[ ${ C }_{ 0 }^{ 2 }$ + ${ C }_{ 1 }^{ 2 }$ + ${ C }_{ 2 }^{ 2 }$ + ........... ${ C }_{ n }^{ 2 }$ ] + 2[ $\underset { 0\le i<j\le n }{ \sum { \sum { { C }_{ i }{ C }_{ j } } } }$ ] $\Rightarrow$ ${ ({ 2 }^{ n } })^{ 2 }$ = $_{ }^{ 2n }{ { C }_{ n }^{ } }$ + 2[ $\underset { 0\le i<j\le n }{ \sum { \sum { { C }_{ i }{ C }_{ j } } } }$ ] $\Rightarrow$ 2[ $\underset { 0\le i<j\le n }{ \sum { \sum { { C }_{ i }{ C }_{ j } } } }$ ] = ${ 2 }^{ 2n }$ - $\frac { (2n)! }{ (n!)(n!) }$ $\Rightarrow$ [ $\underset { 0\le i<j\le n }{ \sum { \sum { { C }_{ i }{ C }_{ j } } } }$ ] = ${ 2 }^{ 2n-1 }$ - $\frac { (2n)! }{ 2(n!)(n!) }$

great solution though

space sizzlers - 4 years, 4 months ago

can you visualise this summation in terms of standard combinatorics ?

Ujjwal Mani Tripathi - 4 years, 5 months ago

Double summation

1 solution

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