
September 11th, 2004, 06:18 AM
#1
How to calculate the number of combinations?
Hello, I have a question here and need your help. I have 3 numbers, each must be in the range of 0 to 38. I need to find how many possible combinations are there...For example, I can have:
0 0 38
38 38 38
1 1 1
2 2 2
.
.
.

September 11th, 2004, 06:30 AM
#2
Permutations and combinations are your best friends.
Since order doesn't matter, meaning you can have (0,0,1) and (1,0,0), you would use permutations.
38 P 3 = 38! / (38  3)! = 38!/35! = 50,616 different combinations you can have.

September 11th, 2004, 12:12 PM
#3
GW is right, it is a question of factorials (the "!" symbol) where, for example, 4! is:
1x2x3x4
Cheers

September 11th, 2004, 01:29 PM
#4
Since order doesn't matter, meaning you can have (0,0,1) and (1,0,0), you would use permutations.
38 P 3 = 38! / (38  3)! = 38!/35! = 50,616 different combinations you can have.
In permutations, order does matter...
Your formula is wrong, imo, because there are 39 possibilities for each number, not 38 (0 to 38), and a permutation doesn't make much sense here (and you're using the term combination to indicate the outcome of a permutation ).
You have 3 numbers, each number needs to be in the range of 0 to 38; the answer is there are 39 to the power of 3 possibilities = 59,319 different combinations, simple as that.
What you're saying with your formula, GW, is that for the first number you have 38 possibilities, for second you only have 37 and for the third only 36. Not only do you have 39 possibilities for the first number, but you also have the same 39 possibilities for the second and the third number (otherwise something like (1,1,1) wouldn't be possible): 39 to the power of 3.
If you really want to make if difficult and assume that ie. (0,0,1) is the same as (0,1,0), you'd use something completely different...
We have our 39 to the power of 3 possibilities, but only 1/3 of them are "valid" since 2/3 of them are redundant:
0 0 1
0 1 0
1 0 0
are all the same  3 possibilities, 2 of them redundant. That would make our formula
(39 to the power of 3) / 3.
From the way the question is asked, though, I'd say you simply have 39 to the power of 3 possibilities...

September 11th, 2004, 02:02 PM
#5
I guess that's one of the + sides of being mathgirl's slave.
When death sleeps it dreams of you...

September 11th, 2004, 02:16 PM
#6
What you're saying with your formula, GW, is that for the first number you have 38 possibilities, for second you only have 37 and for the third only 36. Not only do you have 39 possibilities for the first number, but you also have the same 39 possibilities for the second and the third number (otherwise something like (1,1,1) wouldn't be possible): 39 to the power of 3.
You are right there Neg~.................it is all coming back to me now
I think that the issue is one of "replacement" perhaps Mathgirl could confirm?
As I recall, our national lottery has 49 numbers (149) but there is no replacement, so the solution is 49!
You are quite right about the 038 being 39 numbers...............I should have spotted that
AFAIK in a combination, order does not matter, but in a permutation it does..........or is it the other way round?
Amazing how quickly you forget stuff.............then, it was over 30 years ago

September 11th, 2004, 02:41 PM
#7
I don't know if it's officially called replacement (Mathgirl hates statistics and is asleep... heh), but that's basically what it is, yups.
Lotteries are interesting: if you have 49 lottery balls, and 6 numbers are drawn, you would have 49 possibilities for the first ball, 48 for the second,... Your chance would be:
1 chance in 49*48*47*46*45*44 = 49!/43! = 10,068,347,520.
This is the formula used by GW, but that formula doesn't have anything to do with order. This chance doesn't say that the order doesn't matter, it only says that there is no replacement.
If the drawn balls would be put back in the thingy each time ("replacement"), you would have
1 chance in 49*49*49*49*49*49 = 13,841,287,201.

September 11th, 2004, 03:28 PM
#8
Lotteries are interesting: if you have 49 lottery balls, and 6 numbers are drawn, you would have 49 possibilities for the first ball, 48 for the second,... Your chance would be:
1 chance in 49*48*47*46*45*44 = 49!/43! = 10,068,347,520.
Then throw the power ball in there (+ *49) and the number is even more extream.
\"Life should NOT be a journey to the grave with the intention of arriving safely in an attractive and well preserved body, but rather to skid in sideways, Champagne in one hand  strawberries in the other, body thoroughly used up, totally worn out and screaming WOO HOO  What a Ride!\"
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