Question

Value of "n choose k"
when 0 12 years ago

Answer 1

this is impossible....n must be greater or equal to k

Answer 2

if value must be given it would be 0

Answer 3

And how you come to the conclusion that it is impossible? does the definition of n choose k say so?

Answer 4

not sure mathematically

but think about it practically...if you only start with 3 things, then how will you choose a group of 5 from the 3 things?

Answer 5

http://en.wikipedia.org/wiki/Combination

Answer 6

well, to my knowledge
"n choose k" is not always equal to "choose k for n".
That is only a sub part of using n choose k.
IDK, maybe I'm wrong.
But my textbook proves the existence of that, but I cant understand.

Answer 7

My lecturers famous quote- "Never believe wikipedia definitions" :(

Answer 8

http://math.stackexchange.com/questions/267644/n-choose-k-where-n-is-less-than-k

Answer 9

lol right thats why i said "not sure mathematically"

this is higher math stuff, i usually dont spend too much time thinking about these things

Answer 10

okay, Thank you very much for your help though

Answer 11

if 0 < n < k, then (n choose k) = 0

a simple demonstration is, if you have 3 objects, how many ways can you select 5? well, the answer is 0.

mathematically speaking
(n choose k) = n! / [ (n-k)! k! ]

since n < k, then n - k < 0
(n-k)! = (n-k) (n-k-1) (n-k-2) (n-k-3) ....
loosely speaking (n-k)! is infinitely big.

so [n! / (n - k)! k!] = (n!/k!) (1/(n-k)!) = 0

if n > 0, k < 0, then (n choose k) = 0 by the same argument

the interesting case is when n < 0, and k > 0

then (n choose k) = n (n-1) (n-2) ... (n - (k - 1)) / k!

example:
(2 choose - 5) = (-2)(-3)(-4)(-5)(-6) / 5! = -6

Answer 12

well, at least that would my thought anyways, though the case n <0, and k>0 shows up in binomial series representation

Answer 13

consider Pascals triangle

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1

each number is the sum of the 2 above it .... where do we get the 1s?


0 1 0
0 1 1 0
0 1 2 1 0
0 1 3 3 1 0
0 1 4 6 4 1 0


and where do we get the zeros?


0 0 1 0 0
0 0 1 1 0 0
0 0 1 2 1 0 0
0 0 1 3 3 1 0 0
0 0 1 4 6 4 1 0 0

so the solution with regards to Pascal is that 4 choose 7 is equal to 0, and also stuff like 4 choose -3 is 0 as well