TargetIIT

[b](1) No. of solution of the equation [im]http://codecogs.izyba.com/gif.latex?C%282n%2Cn%29%3D163[/im]
[im]http://codecogs.izyba.com/gif.latex?C%282n%2Cn%29%3D%5Cfrac%7B%282n%29%21%7D%7Bn%21*n%21%7D[/im]
ans= 0  ,1  2  , Infinity

(2)No. of (+ve) Integral solution (x,y) of [im]http://codecogs.izyba.com/gif.latex?%5Cfrac%7B%286x%29%21%7D%7B%28x%21%29%5E%7B6%7D%7D%3Dy[/im] is eual to
ans=0  ,1  , 2 , infinity  [/b]

Q1. zero?

try to prove that [p]2n[/p]C[ss]n[/ss] is even

2) infinity..

coz, wen we put values of x as 1, 2, 3, 4,.....
for each value we get an integral value of y..

as the given expression is always an integer...

subhomoy, For the 2nd part, your answer is ok.. but i dont think you can put this forever and see :D


what you should rather do is to try and see that this is the number of ways to do something.. (can you recognize that?)


Or otherwise you can write the given expression as a product of a few integers.... (nCr is an integer)

For the first part, there is another (not so good method) but may help at other places.... (If the number on the RHS was even!)



What you can look at is the series summation of 2nCn = (nC0)[p]2[/p]+ (nC1)[p]2[/p] +  (nC2)[p]2[/p] +...+  (nCn)[p]2[/p]


Which is definitely greater than 2n[p]2[/p]


Then try to use some more....


You can also try to give some other lower bound on the number 2nCn