Mathmatical Question

[Smiler.]

How does one calculate the number of possible permutations of the following:

a pegboard has 10 holes
a peg can be placed in any or all of the holes
what is the maximum number of combinations of pegs in holes?


Is this statistics??

TIA.

[Neil_H]

Read up on permutations...

[Stevenj214]

Would it not simply be (2^10)-1?

Or if placing no pegs is an option, simply 2^10

Edited by Stevenj214 on Monday 25th May 14:02

[cs02rm0]

Depends how many pegs you have?!

[Famous Graham]

mmertens said:

Smiler. said:

How does one calculate the number of possible permutations of the following:

a pegboard has 10 holes
a peg can be placed in any or all of the holes
what is the maximum number of combinations of pegs in holes?


Is this statistics??

TIA.

10!, I think?

[ewenm]

The OP needs to give us more details - limited number of pegs? Do all holes have to be filled? Is Peg A in hole 1 different to peg B in hole 1?

[mmertens]

Famous Graham said:

mmertens said:

Smiler. said:

How does one calculate the number of possible permutations of the following:

a pegboard has 10 holes
a peg can be placed in any or all of the holes
what is the maximum number of combinations of pegs in holes?


Is this statistics??

TIA.

10!, I think?

You are right. I talk b*ll*x. I'm sorry. 2^10 is right....

[Smiler.]

cs02rm0 said:

Depends how many pegs you have?!

10

[hugo a gogo]

more or less combinations than this, which has 1296?

[dpbird90]

If all 10 pegs are classed as "different", then its the factorial of 10 (write in 10! on a scientific calculator, 1x2x3x4x5x6x7x8x9x10 on old school calculators). If some of the pegs are classed as "the same" then it's a bit different.

Let's say each peg has a different coloured ring around the top of it, except for two which have the same colour (makes it easier to understand). You would then still use the 10! but divide it by the multiple of 2! and 8! (8 being 10-2) so that no. of combinations would be 10! / 2! x 8!

I remember doing this at GCSE but with combinations of letters in people's names. Quite simple if you've got the ! button on your calculator.

[CanAm]

I'm no mathematician, but surely it's Binary. Every combination from 0000000000 to 1111111111 = 1024.
Alan

[Snoggledog]

10 x 10 x 10 iirc

[sstein]

Smiler. said:

How does one calculate the number of possible permutations of the following:

a pegboard has 10 holes
a peg can be placed in any or all of the holes
what is the maximum number of combinations of pegs in holes?


Is this statistics??

TIA.

From reading your description, it's just binary.

So 10 bits, each bit can be 1 or 0 (peg or no peg)

so 2 ^ 10 = 1024 combinations.

So that would be all combinations of pegs in holes from

0000000000 - 1111111111

It gets more complicated if the pegs are different and you want combinations of certain pegs etc.

[Smiler.]

I don't think it's any of the above.

I tried using 5 instead of 10 & came up with 32 variations between all five holes with plugs & any combination down to no plugs & all holes empty.

I broke it down to 6 groups each having 5 combinations + 1 for all 5 + 1 for none.

Using the same logic, I estimate 112 variations using 10, but if anyone knows better, feel free to pitch in.

Ta

[sstein]

Smiler. said:

I don't think it's any of the above.

I tried using 5 instead of 10 & came up with 32 variations between all five holes with plugs & any combination down to no plugs & all holes empty.

I broke it down to 6 groups each having 5 combinations + 1 for all 5 + 1 for none.

Using the same logic, I estimate 112 variations using 10, but if anyone knows better, feel free to pitch in.

Ta

You aren't explaining the problem well enough.

You have 5 holes and you want combinations of pegs in holes and pegs not in holes. Thats 2 ^ 5 = 32 combinations.

If you have the same 10 holes, all combinations are 2 ^ 10 = 1024

Heres, you exmaple with 5 holes

0 0 0 0 0
0 0 0 0 1
0 0 0 1 0
0 0 0 1 1
0 0 1 0 0
0 0 1 0 1
0 0 1 1 0
0 0 1 1 1
0 1 0 0 0
0 1 0 0 1
0 1 0 1 0
0 1 0 1 1
0 1 1 0 0
0 1 1 0 1
0 1 1 1 0
0 1 1 1 1
1 0 0 0 0
1 0 0 0 1
1 0 0 1 0
1 0 0 1 1
1 0 1 0 0
1 0 1 0 1
1 0 1 1 0
1 0 1 1 1
1 1 0 0 0
1 1 0 0 1
1 1 0 1 0
1 1 0 1 1
1 1 1 0 0
1 1 1 0 1
1 1 1 1 0
1 1 1 1 1


Do the same with 10 and you will have 1024 combinations.

[ewenm]

Unless the pegs are different, so peg A in hole 1 is not the same as peg B in hole 1.

[sstein]

ewenm said:

Unless the pegs are different, so peg A in hole 1 is not the same as peg B in hole 1.

But he has already said using 5 holes = 32 combinations then its reasonable to assume that he isn't including that.

So if 5 = 32
10 = 1024

[Penny-lope]

Who honestly cares?

[sstein]

There's your 1024 combinations of pegs in holes with 10 holes. From all empty to all full.

http://www.speed-shot.co.uk/myfile2.txt

[ThatPhilBrettGuy]

Penny-lope said:

Who honestly cares?

There's an ironic post.