Question

Binomial question:
simplify :
C(50,0) * C(50,1) + C(50,1) * C(50,2) .....+C(50,49) * C(50,50)

C(n,r) has the usual meaning i.e. n(combination)r

Answer 1

there are 4 options, though C(50,25) is an option but the ans given at the back is
C(100,51)

Answer 2

sorry i actually need to prove it

it should be C(2n,n-1) or C(2n,n+1) hence answer is correct ..

Answer 3

first we use (1+x)^n (x+1)^n =(1+x)^(2n)
i am actually very lazy in writing ...

can u check that the LHS of ur question is just the coefficient of x^(n-1)
hence from RHS we need the coefficient of x^(n-1) which is C(2n,n-1)
here n=50
hence C(100,49) =C(100,100-49)=C(110,51)

Answer 4

a more genaralised result is
C(50,0) * C(50,r) + C(50,1) * C(50,r+1) .....+C(50,n-r) * C(50,50) = C(2*50,50-r)
though it can be generalised further
but u can proceed as above...

Answer 5

i shall get back to you,,please gimme some time..

Answer 6

if possible sure ..

Answer 7

as i hardly remain online...

Answer 8

well,,i tried but i couldnt understand the concept really sir,,but i rote the formulla ! :P

next ques was C(50,0) ^2 + C(50,1) ^2. .....(C(50,50) ^2
so according to the formulla,,r=0 and on plugging in that, we get the right ans i.e. C(100,50)

but what about this kind ques :
C(100,0)* C(200,150) + C(100,1) * C(200,151) ..... C(100,50) * C(200,200) ?

Answer 9

here u take (1+x)^100 *(x+1)^200 =(1+x)^300

Answer 10

okay,,following..

Answer 11

sir ? hmm..?

Answer 12

is LHS the coefficient of x^50 then from RHS u need the coefficient of x^50 ( if i am wrong just find the proper power of x)
hence coefficient of X^50 in RHS is C(300,50) or C(300,150)
is the ans correct

Answer 13

nCr = nC(n-r)

Answer 14

C(50,0) * C(50,1) + C(50,1) * C(50,2) .....+C(50,49) * C(50,50)

C(50,0) * C(50,49) + C(50,1) * C(50,48) .....+C(50,49) * C(50,0)
http://en.wikipedia.org/wiki/Vandermonde's_identity

Answer 15

C(50,0) * C(50,1) = C(50,49) * C(50,50)
C(50,1) * C(50,2) = C(50,48) * C(50,49)
C(50,2) * C(50,3) = C(50,47) * C(50,48) etc

Answer 16

let \( r=49\)
C(50,0) = C(50,50)
C(50,1) = C(50,49)
C(50,2) = C(50,48)
...
the expression is like
\[ \sum_{k=0}^{50}\binom{50}{k} \binom{50}{49-k}\]
you can directly apply Vandemonde's identity to get
\[ \binom{100}{49} \text{ or } \binom{100}{51}\]

Answer 17

Woops
\[ \sum_{k=0}^{49}\binom{50}{k} \binom{50}{49-k} \]

Answer 18

thank you :)