plz solve it

[chintz]

if nC14 = nC16 ; then 32Cn = ???

[ecodude]

nC14 = n! / (n-14)! * 14! = nC16 = n! / (n-16)! * 16!
=> 1/{(n-14) *(n-15) * (n-16)!* 14!} = 1/{ (n-16)! * 16*15*14! }
=> (n-14)(n-15) = 16 * 15 => n^2 - 29n -30 =0 => n=30 or n= -1.
Since n can't be -1 , it is 30. => 32C30 = 496 .
Tell me if i made some error

[Dkumar]

I got the same answer dude.

[arjmen]

General Case:
nCr = nC(n-r) since number of ways of picking 'r' of 'n' things is the same as the number of ways of not picking 'n-r' of 'n' things
n = r+(n-r)

In the question, n = 14+16 = 30

32C30 = 32!/(30!2!) = 16*31 = 496

[ecodude]

good concepts arjmen. feels good to see a better method.

[chintz]

good method to crack arjmen.

[babra]

y do u have 2! in the denominator (nCr= nPr/r!)........ por favor, El Señor explain

[chintz]

v can write nCn-r instead of nCr

so 32C30 = 32C32-30 = 32C2 = 32*31/2 = 496

Got it????

[babra]

yup thank u .....