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...

Question

ganeshie8 · Answer

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

anonymous · Answer

thats 2n choose n

ganeshie8 · Answer

Yes, lets count it in an alternative way

ganeshie8 · Answer

how many committees will be there with out women ?

anonymous · Answer

*

anonymous · Answer

1

ganeshie8 · Answer

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

anonymous · Answer

n choose n-1

ganeshie8 · Answer

try again

anonymous · Answer

n choose n-1 multiplied by n?

ganeshie8 · Answer

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 · Answer

does that make sense

anonymous · Answer

yes, i get it

anonymous · Answer

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

ganeshie8 · Answer

yes find it, after that you will see the pattern

anonymous · Answer

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

ganeshie8 · Answer

take your time, we're almost done!

anonymous · Answer

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

anonymous · Answer

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

ganeshie8 · Answer

Yep!

anonymous · Answer

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

anonymous · Answer

oh wait nevermind

anonymous · Answer

i get it

ganeshie8 · Answer

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

anonymous · Answer

sure :D

ganeshie8 · Answer

take a screenshot and attach if psble

anonymous · Answer

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

ganeshie8 · Answer

take ur time

anonymous · Answer

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 · Answer

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

anonymous · Answer

yes im just typing it up now

anonymous · Answer

sorry the codeisnt working...

ganeshie8 · Answer

make this correction :  $k\le n$

ganeshie8 · Answer

other than that, the proof looks good!

anonymous · Answer

ok 
thanks!

anonymous · Answer

can you help me with the second part please?

anonymous · Answer

hello? are you still here @ganeshie8

ganeshie8 · Answer

Hey!

anonymous · Answer

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

ganeshie8 · Answer

Consider a $n\times n$grid
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anonymous · Answer

right

anonymous · Answer

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