MHB Proving $\sum_{k=0}^{n} \binom{n}{k}^2 =\dfrac{(2n)!}{(n!)^2}$

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The discussion centers on proving the identity $\sum_{k=0}^{n} \binom{n}{k}^2 = \dfrac{(2n)!}{(n!)^2}$. Participants share various approaches to demonstrate this combinatorial identity, highlighting its significance in combinatorics. The proof often involves interpreting the left side as counting paths in a grid or using generating functions. Several users express appreciation for elegant solutions and insights shared throughout the discussion. This identity is a well-known result in combinatorial mathematics, illustrating the relationship between binomial coefficients and factorials.
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prove:
$\sum_{k=0}^{n} \binom{n}{k}^2
=\dfrac{(2n)!}{(n!)^2}$
 
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Albert said:
prove:
$\sum_{k=0}^{n} \binom{n}{k}^2
=\dfrac{(2n)!}{(n!)^2}$

out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

now let us make 2n objects into 2 groups of n objects each.

for picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

the 2 above are same as it shows the number of ways in 2 different ways

so
$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result
 
kaliprasad said:
out of 2n objects n objects can be chosen in

$\dfrac{(2n)!}{(n!)^2}$ ways

now let us make 2n objects into 2 groups of n objects each.

for picking n objects from the set we need k objects from 1st set and n-k from 2nd set and n varies from 0 to n so number of ways

$\sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

the 2 above are same as it shows the number of ways in 2 different ways

so
$\dfrac{(2n)!}{(n!)^2}= \sum_{k=0}^{n} \binom{n}{k} \binom{n}{n-k}$

now as $\binom{n}{k} = \binom{n}{n-k}$ so we get the result
nice solution !
 
Thread 'erroneously finding discrepancy in transpose rule'

Thread 'erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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