- anonymous

I need help to prove that $$\binom{n}{0}^2 + \binom{n}{1}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}.$$ using committee forming...

- schrodinger

See more answers at brainly.com

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- ganeshie8

suppose there are \(n\) men and \(n\) women
and you want to choose a committee consisting of \(n\) people

- anonymous

thats 2n choose n

- ganeshie8

Yes, lets count it in an alternative way

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- ganeshie8

how many committees will be there with out women ?

- anonymous

*

- anonymous

1

- ganeshie8

Yes, how many committees will be there with exactly 1 women ?

- anonymous

n choose n-1

- ganeshie8

try again

- anonymous

n choose n-1 multiplied by n?

- ganeshie8

you can choose \(1\) women from the group of \(n\) women in \(\binom{n}{1}\) ways
after that, the remaining \(n-1\) men can be chosen from the group of \(n\) men in \(\binom{n}{n-1}\) ways
so total \(n\) member committees with exactly \(1\) women is \(\binom{n}{1}*\binom{n}{n-1}\)

- ganeshie8

does that make sense

- anonymous

yes, i get it

- anonymous

now would i find the number of ways to make a committee with two women?

- ganeshie8

yes find it, after that you will see the pattern

- anonymous

alright, ill get back to you with the results :D

- ganeshie8

take your time, we're almost done!

- anonymous

hang on... n choose k and n choose (n-k) give the same result

- anonymous

so the total possible combinations is just sums of the squares....

- ganeshie8

Yep!

- anonymous

but how do i equate it to 2n choose n?

- anonymous

oh wait nevermind

- anonymous

i get it

- ganeshie8

good :) id like to see the complete proof if you don't mind

- anonymous

sure :D

- ganeshie8

take a screenshot and attach if psble

- anonymous

i need to go eat dinner, i'll send it to you in about an hour, is that ok?

- ganeshie8

take ur time

- anonymous

i also need to prove the same thing using the "block walking" method... but i dont know how. Do you think you can try to help me with this too? please?

- ganeshie8

sure that is also an interesting way to count
first, may i see the previous proof...

- anonymous

yes im just typing it up now

- anonymous

sorry the codeisnt working...

- ganeshie8

make this correction : \(k\le n\)

- ganeshie8

other than that, the proof looks good!

- anonymous

ok
thanks!

- anonymous

can you help me with the second part please?

- anonymous

hello? are you still here @ganeshie8

- ganeshie8

Hey!

- anonymous

hi! can you please help me prove this with the "block walking"

- ganeshie8

Consider a \(n\times n\)grid
|dw:1440165928386:dw|

- anonymous

right

- anonymous

number of paths from bottom left to top right = \(\binom{2n}{n}\)
|dw:1440168478785:dw|

- anonymous

suppose your friend's home is located at \((k,~n-k)\), where \(k\in \left\{0,1,2,\ldots ,n \right\}\).
then number of paths through \((k,n-k)\) is given by \(\binom{n}{k}*\binom{n}{n-k} = \binom{n}{k}^2\)
.....
https://proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients

Looking for something else?

Not the answer you are looking for? Search for more explanations.