Why is there not a simpler way to do this?
>>8066222
Looks pretty simple to me
>>8066222
Pascal's triangle
>>8066222
wow that notation [math]{}_n C_k [/math] is actually pretty good.
The American notation is shitty and the French Pascal notation is usually confusing.
It is simple usually the (n-r)! would cancel out the majority of n!. Or if r! was big enough that could cancel out the majority of the n!. Your answer always results into a whole number.
>>8066243
Can you post these notations?
>>8066253
american notation [math] { k \choose{n} } [/math] : completely fucks up your tex, ugly
french notation [math]C^k_n [/math] : streamlined and pretty but too many confusions with exponents, Catalan numbers, constants...
Notation [math] {}_nC_k [/math] not so bad but in fact it's still confusing :(
Also the identity [math]C^k_n = \frac{n!}{k!(n-k)!}[/math] is just basic counting : from n students, first choose the first, the second, then... then kth (that is, the k best students in order)
You've got [math]n(n-1)...(n-k+1) = n!/(n-k)![/math] possible choices for the ordered k best students, and there are [math]k![/math] possible orderings, sooooo....
>>8066283
How about [math]_nC^k[/math]?
>not [math] \text{(choose n k)} [/math]
>>8068336
>fckn MathJax
I fckn give up, you win MathJax
funny enough, I was wonderug abiut this yesterday.
Maybe instead of expressing it in terms of the factorial/Gamma, is there a way to express C that makes it evident that it's in N and not just Q?
>>8068431
I would also like to know this....
>>8066222
In Australia we use [math]^nC_r[/math]
[math]^nC_r[/math] master race
>>8066283
how about C(n,k). Fuck aesthetics.
>>8068568
Could be a seen as a multivariable function like [math] F(x,y) [/math]
>>8068573
..it is a multivariable function.
>>8066283
How fucking underage are you where you've never seen [math]_nC_k[/math] before?
>>8068901
Haven't seen it either.
>>8068573
C(n;k)
AYYYYYYY
>>8066283
nCk is the american notation for combinations. The "american notation" you referenced is the universal notation for binomial coefficients (which happen to be the same thing)
>>8068573
so? kill yourself retard
>>8066222
Because the number of subsets of a (finite) set is a slightly tricky thing to work out, requiring 2 or 3 pieces of information in your formula/algorithm. Thus, what you have presented is about as simple as it gets. Any simpler and it'd be wrong.
Two things worth knowing cold:
1) Where n is always >= k, Pascal's triangle can be rewritten (0,0) ... (1,0) , (1,1) ... (2,0), (2,1), (2,2)... The zero thingy starts on the tippy top, and the k-zeroes continue off down and to the left (or right, it's symmetric after all).
2) A given row of the triangle provides the answer to the first clause of this post for some given n, in the cases from k = 0... up. and the answer is always 2^n (t. Cantor). But the utility of the triangle here is that it reminds us that the cases are always symmetric, and it enumerates the cases a priori for some given k.
As long as you keep the above two chunks of information in mind, you can make sense of combinations from scratch, in my experience. And, like I said earlier, the original question is a slightly tricky business that requires you to keep track of 2-3 pieces of information. You could either use the ones in the OP formula, or you could start from the chunks I've suggested and work backwards, always (re)-discovering that pascal's triangle "just works" for the right cases, case-by-case.
>>8068431
>>8068528
If you use the recursive formula, then [math]C(n,k)[/math] ends up as sum of only two different but repeated terms, namely [math]C(m,m)[/math] and [math]C(n-k,0)[/math], which are each equal to [math] 1 [/math].
In this recursive representation, it's a sum of [math] 1 [/math]'s and thus an integer.
For example
[math] C(4,3) = \dfrac{4!}{3!(4-3)!} = \dfrac{4!}{3!} = 4 [/math]
or
[math] C(4,3) [/math]
[math] = C(3,2) + C(3,3) [/math]
[math] = [C(2,2) + C(2,1)] + C(3,3) [/math]
[math] = [C(2,2) + [C(1,1) + C(1,0)]] + C(3,3) [/math]
[math] = 4 [/math]
>>8069158
I just started playing games with more linear such schemes (not that I learned something, but maybe someone is interested)
f[m_, 0] = a[m];
f[0, m_] = b[m];
f[m_, m_] = c[m];
f[n_, k_] := A*f[n, k - 1] + B*f[n - 1, k] + C*f[n - 1, k - 1]
Table[{n, k, f[n, k]}, {n, 1, 4}, {k, 1, 4}]
>>8069132
this triggers me
