I'm not sure what you are asking here. Could you clarify?

[Merlin3189]

Edit = Solved. My error.
1- $\Big( \frac {n_1 +n_2}{n_1} \Big)$ in Siegel,S "Nonparametric Statistics" P.153 re. Randomization test for 2 independant samples.
eg. $\Big( \frac {4 + 5}{4} \Big) = 126$
Used in calculating number of "equally likely ways" data could have been assigned under H₀
The brackets seem to be in heavier type and larger than BODMAS brackets.

So what does this notation mean? How do I evaluate it?

I should say I am not a mathematician, though I do have access to a retired A-level teacher, who has not been able to enlighten me. I do have A-level maths (but Pure with Mechanics, rather than Stats) and have used Siegel as a recipe book for doing stats in behavioural science. I don't remember using this test before, but its description seems appropriate to my data. It's simply that I can't evaluate his formula, because I don't recognise the notation.

2-Relevant ⁿC_r = $\frac{n!}{r! (n-r)!}$
eg. ⁵C₄=$\frac{5!}{4!1!}$=5

ⁿP_r = $\frac{n!}{ (n-r)!}$
eg. ⁵P₄ = $\frac{5!}{ (1!}$ = 120

Binomial Coefft = $\frac{n!}{k! (n-k)!}$
which looks remarkably like ⁿC_k

n! = n(n-1)(n-2)...1 eg. 5!=5x4x3x2x1 = 120

rⁿ = r*r*r*r... n times eg. 5⁴=5x5x5x5=625

3-attempts
Searched Google for Math notation re. brackets and braces.
Found examples of its use for Combinations and for Binomial Coeffts.
Evaluated
⁵C₄=$\frac{5!}{4!1!}$=5
⁹C₄=$\frac{9!}{4!1!}$= 9x8x7x6x5 > 126 EDIT - Error!
⁹C₅=$\frac{9!}{5!1!}$= 9x8x7x6 > 126 EDIT - Error!
Could not work out a different result for Bin(4+5,4 or 5) because it just seemed to be the same as the combinations.

Factorised 126 = 2 x 3 x 3 x7 (maybe = 3! x (3-2)! x 7! / 6! ?)

I suppose I could work out the number of ways I think the data (9 items in the example) could be assigned to 2 groups (of 4 and 5), but I suspect I'd arrive at
ⁿC_r again.

In the group of 5, there are 9 choices for 1st position, 8 for next, ..
= 9x8x7x6 = 3024
Once these are assigned, all the rest must be in the group of 5.
--------------------
There are 9 data, each can be in one of two groups, so
2 choices for first, 2 choices for second, ...
= 2x2x2...= 2^9= 512 ways of assigning 9 data to 2 groups, from AAAAAAAAA to BBBBBBBBB via for eg. AABBABABB
But only 4 can be in one group and 5 in the other, so I should reduce this.
I need to count only AAAABBBBB , BABABABAB, etc and discard AAAAABBBB, AAAAAAAAA, AAAAAAAAB etc.
So I can discard
1 like AAAAAAAAA and 1 like BBBBBBBBB
9 like ABBBBBBBB and 9 like BAAAAAAAA
36 like AABBBBBBB and 36 like BBAAAAAAA
84 like AAABBBBBB and 84 like BBBAAAAAA
and 126 like BBBBAAAAA

Total 386, leaving 512-386= 126 as required

So I'm happy that he got the correct result, but how did he calculate it from (5+4)/5 or from (5+4)/4 ?

I ended up using 2^(5+4) - 2xC(9,0) -2xC(9,1) -2xC(9,2) -2x(9,3) -2xC(9,4) -C(9,5) which is cumbersome to say the least.
Since he uses this weird notation, it seems likely that it is a well known function.

And since when I come to apply it, I have 65 data in groups of 17 and 48, my ab initio process will be tedious! If I understood his method (hidden in this bracket notation) perhaps it would help me see something I have not observed.
I understand that if, I want to be a mathematician, I must work out everything from first axioms and build my own framework, but I'm just trying to evaluate my data. Much as I'd love to be a mathematician, I don't think I have the time.


1. Homework Statement

Homework Equations

The Attempt at a Solution

 

[PeroK]

You've simply got a non-standard notation for binomial coefficients.

I learned these as $^nC_r$ but $\binom{n}{r}$ seems to be more common.

$\binom{5+4}{4} = \binom{9}{4} = 126$

Rathers gives it away.

There's more here:

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

 

[Merlin3189]

Thanks for the response. I'm sorry, I just miscalculated when I evaluated it, or I would have realized that's what it meant.
I wish I could delete it!