Maths help please

[jonnie5]

I'm having a bit of an 'empty moment' and wondered if the power of PH could help.

If I have 6 options and I can choose any number of them, how many possible combinations are there? (assuming you choose at least one)

I have a number of answers, I'm damned if my head is going to work today! :irked:

TIA
Jon

[Jinx]

jonnie5 said:

I'm having a bit of an 'empty moment' and wondered if the power of PH could help. If I have 6 options and I can choose any number of them, how many possible combinations are there? (assuming you choose at least one) I have a number of answers, I'm damned if my head is going to work today! TIA Jon

The answer is 6! or 6*5*4*3*2*1

[samwilliams]

If I understood what you're asking properly, then I think there are 63 different combinations. But then again, I may be wrong

Sam

[Bitter'n'twisted]

Yes, aree with above, should be 63.

>> Edited by Bitter'n'twisted on Wednesday 13th July 16:56

[jonnie5]

Jinx said:

The answer is 6! or 6*5*4*3*2*1

I started with 6x5x4x3x2x1 which equals 720, just seems a lot..

samwilliams said:

If I understood what you're asking properly, then I think there are 63 different combinations. But then again, I may be wrong Sam

To clarify, what I'm asking is if I have 6 options and I can have any number of them, what is the total number of possible permatations of this.
e.g. it could be option1 and option3, or option1, option 2 and option 4, or all of them, or just one....

63 sounds more like what I was expecting, how do you get there?

Thanks

[Jinx]

Ok miss understood the question - the number or permutations = 720 (order relevent)

[ledger]

does the order matter, i.e is option 1, option 2 the same selection as option 2, option 1.

It's permutations and combinations see :-)

[jonnie5]

Jinx said:

Ok miss understood the question - the number or permutations = 720 (order relevent)

OK, the order is not relevant. Imagine them as options on a car (see what I did there..)

ledger said:

does the order matter, i.e is option 1, option 2 the same selection as option 2, option 1. It's permutations and combinations see :-)

Yes, to confirm they are the same selection


>> Edited by jonnie5 on Wednesday 13th July 17:05

[samwilliams]

The way I got there was using the little combinations formula thingy nCr (vaguely remember it from maths lessons)

Only pick one: 6C1 or 6!/(1!x5!) = 6
Pick two: 6C2 or 6!/(2!x4!)=15
3: 6C3=20
4: 6C4=15
5: 6C5=6
6: 6C6=1

So, the total is 63.

[ledger]

jonnie5 said:

Jinx said: Ok miss understood the question - the number or permutations = 720 (order relevent) OK, the order is not relevant. Imagine them as options on a car (see what I did there..)

right it is combinations then (in case you wanted to look it up in a maths text book)

the formula for the number of choices of "c" from a total choice of "n" (referred to as "n choose c" is
n!/[c!*(n-c)!]
where ! is the factorial operator (i.e. 4! = 4*3*2*1)

so for your problem we have
6 choose 1 + 6 choose 2 + 6 choose 3 + 6 choose 4 + 6 choose 5 + 6 choose 6

= 6 + 15 + 20 + 15 + 6 + 1 = 63


god dam fat fingers, beaten to it again





>> Edited by ledger on Wednesday 13th July 17:11

[jonnie5]

Thank you all

It seems so simple when you spell it out like that.

Cheers
Jon






right, better get back to work then. At least I can demonstrate my mathmatical genius to my client tomorrow!

[Bitter'n'twisted]

jonnie5 said:

63 sounds more like what I was expecting, how do you get there? Thanks

I think of it as binary digits.
In your case 6 binary digits. 111111
Which in decimal is 1+2+4+8+16+32 = 63.

If it was 8, it would become 255 combinations.

[simpo two]

I'm a bit late here but the key is whether the one you choose is still available for choosing the next time. 'With replacement' rings a bell.

So if you're choosing chocolates and eating them as you go, the number of options is much lower than if it was a combination safe, where, for example, you could have 555555 as a combination.

[Incorrigible]

You have 6 single options

With 2 options you can pick any of the remaining 5 with each one

1+2, 1+3, 1+4, 1+5, 1+6
2+3, 2+4, 2+5, 2+6
3+4, 3+5, 3+6
etc

5 + 4 + 3 + 2 + 1 = 15

with 3 options

1+2+3, 1+2+4, 1+2+5, 1+2+6; 1+3+4, 5, 6; 1+4+5, 6

4+3+2+1
3+2+1
etc

10+6+2+1 = 19

etc

6+15+19+12+10+1=63 or something like that

Or you make it neater factorials

[V8 Archie]

Bitter'n'twisted said:

I think of it as binary digits. In your case 6 binary digits. 111111 Which in decimal is 1+2+4+8+16+32 = 63. If it was 8, it would become 255 combinations.

Almost - for each option you can either have it or not so for 6 options there are 2^6 = 64 options.

The way the question is phrased it appears that we are not allowed to "not select any of them" so it's 2^6 - 1 = 63

BnT's post forgets that 000000 is a valid binary construction.

[foster3jd]

Blimey, talk about making hard work of it... only understood the question once bitter'n'twisted mentioned binary!!!

Can we have the next problem in hexadecimal please!

[david_s]

If you have 6 choices each time then the number of permutations is 6^6, or 46656, I think. it is only the lower number if the number of choices reduces with each choice made. ie if you are choosing from 6 options the first time, then 5 the second and 4 the third etc. If it is a car option choice such as 6 colours followed by six wheel types followed by 6 trim options etc, then it is 6x6x6x6x6x6. But it is a long time since I did much maths so don't trust me.