combination?

Question

using the numbers one to one hundred
how many combinations can you have

you can use any number of the numbers they can't repeat however. ie. (1,3) or (6,4,2,3,)

now suppose each number also was a, b or c
ie. 1a, 1b ,1c
you cannot use the same number twice in the combination ie. you can't use 1a and 1b in the combination
how many combinations now

I need to figure this out for an advanced biology paper.on neural connections.
my expertise is biology and not mathamatics.

thanks for the answers to my other question
now I realize Ishould have asked how many permuations or combinations are possible
can you please just give me a number
PLEASE

Answer 1

The number of permutations of n objects taken r at a time is:

n!/(n-r)!

There is one way to permute no items.
There are 100 ways to permute 1 item.
There are 9,900 (100 x 99) ways to permute 2 items.
There are 970,200 (100 x 99 x 98) ways to permute 3 items.
...
There are 100! ways to permute all 100 items. ********

So allowing all numbers, you have the sum of 0! to 100! or 1! to 100! if you discount zero neurons.


The number of combinations of n objects taken r at a time is:

n!/r!(n-r)!

For 100 elements:
there is one way to combine 0 or 100 elements;
there are 100 ways to combine 1 or 99 elements;
there are 4950 ways to combine 2 or 98 elements...

In fact, Pascal's triangle gives the list of combinations of 0 to 100 elements and they sum to 2^100. If you discount 0 neurons, it is 2^100 - 1.

If you have 3 selections, a, b, c and have to arrange 100 (1a, 1b or 1c; 2a, 2b or 2c; ... ; 100a, 100b or 100c) of them, that's 3^100.

Answer 2

Well...

If you can use any of the numbers from 1 to 100, in any size grouping, but you can't repeat, then you can think of it as, for every number from 1 to 100, you either use it or you don't.

As a result, I think the answer to your first question would be 2^100.

The second one, if I understand correctly, would just be 3 x 2^100. That's assuming that you mean that each number is also assigned a letter a, b, or c, and once assigned, cannot have another letter assigned to it.

PS -- my answer would be different if:

(1, 2, 3, 4) should be considered different than
(2, 3, 4, 1).

Answer 3

Does the sequence depend upon order? If not, then either a number can be there or not- so each additional possible number doubles the number of sequences- for 100 numbers, 2^100 possible sequences. Subtract one if you don't want to include the empty set- where no number is selected.
If order matters- let me think- I think I'd sum the cases that contain a specific number of members- zero members is one case, and can only be obtained one way. 1 member can be obtained 100 ways. 2 members can be obtained 100 choose 2 ways, which is 100!/(2!98!) ways- but you can rearrange the members 2 ways- e.g. 22,6 or 6,22, for a total of 2!*100!/2!98! ways. Okay, for n members, you'll have n!*100!/n!*(100-n)! ways- so you'll sum from i=0 to 100 (possible number of members) 100!/(100-i)!.
Number of unique groups=sum(i=0,100) {100!/(100-i)!}=
100!*sum(i=0,100){1/(100-i)!}=100!*sum(j=0,100){1/i!}
Again, subtract one if you don't want to include the empty case.

For the second part, we'll add a factor of 3^n in the above sum.
Number of unique groups=sum(i=0,100){100!*3^i/(100-i)!}
Subtract one if you don't want the empty case.

Whew- this took a little while- there's probably a very elegant solution sitting just above this now... :>

Answer 4

The number is approximately 100!, which is approximately 9.3E157. That's a BIG number -- far larger than the total number of particles in the universe.

Answer 5

100!