# Notation - Perms & Combs?

1. Mar 4, 2015

### 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 nCr = $\frac{n!}{r! (n-r)!}$
eg. 5C4=$\frac{5!}{4!1!}$=5

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

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

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
nCr 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.
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 wierd 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. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Mar 4, 2015
2. Mar 4, 2015

### 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

3. Mar 4, 2015

### Merlin3189

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