Quick Math Combination Question

FabienO · 28 Dec 2011 at 14:31

Just here at work and stupidly us developers can't find the easiest way to figure out how many combinations of the numbers 1-7 there are, regardless of order.

So 1,2,3 is the same as 3,2,1.

Any quick solutions out there?

My google-fu is weak.

PermaBanned · 28 Dec 2011 at 14:32

It's "maths".

And i think there's an algorithm for this, i vaguely remember it from GCSE. But i can't remember what it's called. I'll have a look for you...

Skillmister · 28 Dec 2011 at 14:35

7x6x5x4x3x2x1 isn't it?

7 ways to choose the first number, 6 to choose the second etc... may be wrong haven't done it in a few years

mufc802 · 28 Dec 2011 at 14:37

If you have a scientific calculator there is an easy way to do this. Type in 7! which is the same as 7x6x5x4x3x2x1. The answer is 5040.

FabienO · 28 Dec 2011 at 14:38

PermaBanned said:
It's "maths".

And i think there's an algorithm for this, i vaguely remember it from GCSE. But i can't remember what it's called. I'll have a look for you...

You need to capitalise your I's, also sentences shouldn't start with and or but.

Appreciate it.

FabienO · 28 Dec 2011 at 14:39

mufc802 said:
If you have a scientific calculator there is an easy way to do this. Type in 7! which is the same as 7x6x5x4x3x2x1. The answer is 5040.

Doesn't that include order? Mine is for without.

Kouba · 28 Dec 2011 at 14:40

FabienO said:
You need to capitalise your I's, also sentences shouldn't start with and or but.

Appreciate it.

Good comeback for the grammar nazi!

TheCenturion · 28 Dec 2011 at 14:41

Is it not just 7C3?

= 35

Skillmister · 28 Dec 2011 at 14:42

TheCenturion said:
Is it not just 7C3?

= 35

Pretty sure he wants to use all the numbers.

Skillmister · 28 Dec 2011 at 14:43

Problem 3. a) How many different arrangements are there of the letters of the word numbers?
7! = 5,040

http://www.themathpage.com/aprecalc/permutations-combinations.htm

TheCenturion · 28 Dec 2011 at 14:44

Skillmister said:
Pretty sure he wants to use all the numbers.

in which case -

7C1 + 7C2 + 7C3 + 7C4 + 7C5 + 7C6 + 7C7

= 127

Or am I completely off track?

What I've done here is to say how many ways can you arrange 7 items 1 item at a time, then 2, then 3, then 4, then 5, then 6, then 7, regardless of order.

FabienO · 28 Dec 2011 at 14:45

Think of it as this.

I have 7 products on my website. I want to create a link for every possible combination that people could buy those.

Hence why order doesn't matter. If I bought products 3, 5 and 7. It's the same as me buying 5, 7 and 3.

So the above is only 1 link. How many links to cover all products?

PermaBanned · 28 Dec 2011 at 14:47

FabienO said:
You need to capitalise your I's, also sentences shouldn't start with and or but.

Appreciate it.

Spelling before grammar

On a calculator it's called the factorial function, the "algorithm" would just be something like x*(x-1)*(x-2)... until you get to one.

Although for some reason my stupid brain thought that you could potentially also do it with bubble sort and the maximum number of passes function...

Skillmister · 28 Dec 2011 at 14:47

In that case i have no idea.

Dr Delphi · 28 Dec 2011 at 14:51

mufc802 said:
If you have a scientific calculator there is an easy way to do this. Type in 7! which is the same as 7x6x5x4x3x2x1. The answer is 5040.

This is correct the factorial of the number of items you want to arrange will give you the possible number of combinations.

Simple set would be 1 and 2 so possible combinations are:
1 2
2 1
Two, thus the factorial of 2 (2!) is 1x2 = 2.

QED

TheCenturion · 28 Dec 2011 at 14:53

We could right them out!

First, all the ones:

1, 2, 3, 4, 5, 6, 7

Now all twos

1 2, 1 3, 1 4, 1 5, 1 6, 1 7,
2 3, 2 4, 2 5, 2 6, 2 7
3 4, 3 5, 3 6, 3 7
4 5, 4 6, 4 7
5 6, 5 7
6 7

... I'm bored now

Greebo · 28 Dec 2011 at 14:54

Dr Delphi said:
This is correct the factorial of the number of items you want to arrange will give you the possible number of combinations.

Simple set would be 1 and 2 so possible combinations are:
1 2
2 1
Two, thus the factorial of 2 (2!) is 1x2 = 2.

QED

Except that in the ops case 1 and 2 is the same as 2 and 1 and his answer would be 1.....................

panyan · 28 Dec 2011 at 14:55

FabienO said:
how many combinations of the numbers 1-7 there are, regardless of order.

So 1,2,3 is the same as 3,2,1.

if it is regardless of order that makes a big difference

mufc802 said:
Type in 7! which is the same as 7x6x5x4x3x2x1. The answer is 5040.

i dont think this is correct as this takes 123 and 321, etc as different numbers

TheCenturion said:
Is it not just 7C3?

= 35

i think this is the correct way as you are looking for "combinations" not "factorials"

TheCenturion · 28 Dec 2011 at 14:57

Okay, 127 must be the right answer, if order doesn't matter.

If you only want one number, then you have 7C1 (ie: 7) to choose from.

If you want two numbers, you have 7C2=21 to choose from.

If you want three numbers, you have 7C3=35 to choose from.

7C4=35
7C5=21
7C6=7
7C7=1

1+7+21+35+35+21+7 = 127

[This assumes that order does not matter. Ie - if you pick number 5 and 2, that's the same as 2 and 5]

Dr Delphi · 28 Dec 2011 at 14:59

Greebo said:
Except that in the ops case 1 and 2 is the same as 2 and 1 and his answer would be 1.....................

Ah comprehension fail. Sorry. Quick google found this:

Combinations

Combinations are unique subsets of objects, ignoring order. The number number of 5 card poker hands that can be dealt from a 52 card deck is a combinations problem. The 120 (i.e. 5!) different arrangements of cards in a hand containing 10,J,Q,K,A of hearts are all the same in terms of combinations. This is true of every other hand, so you can see that there a many fewer combinations than permutations of things. In fact, there are always 1/R! fewer combinations than permutations, so the total combinations for R of N objects is the number of permutations divided by R!, i.e. N! /(N-R)! / R!. For our 2 of 5 example this is 5!/3!/2!=10. For poker hands the total number is 52!/47!/5!=2,598,960.