how many 'six different numbers' combinations can you make w/ 1-54?

Question

would a permuation 'n!' work? dont think so becuase it would take all the numbers...or would i have to divide n!/6 ?

Answer 1

i should know this, I did a semester of statistics....

hmmmm.....

I guess I should have went to class more...... I hope I passed the exam.....

good luck....

..

Answer 2

54 to the sixth power only works if you want six numbers from one to 54 and they don't have to be different numbers. That is, you could have six ones.

Since you want six different numbers, you have to perform the permutation 54P6 (i.e. 54*53*52*51*50*49). This results in 18,595,558,800 order specific permutations. This will take into account all the different possibilities of numbers, however, you don't state whether or not the order matters. For example, are 1-2-3-4-5-6 and 1-3-2-4-5-6 the same or different combinations?

If order is unimportant you would perform the combination function, which is 54P6 divided by 6!. (i.e. 54*53*52*51*50*49/6!). This results in 25,827,165 possible combinations where the order of the numbers is unimportant.

Answer 3

The answer is simpler when using sigma (sum).

n=54 E 6....n(n-1)

Where n is the starting number that you multiply by the following (lower) number. Then multiply the total with the following (lower) number. Then multiply the total with following (lower) number...etc.

n is the first number.
E signifies sigma (supposed to look like an old Z)
6 is the number of numbers.
n(n-1) is the procedure.

After writing the "formula", you do the math. It looks altogether something like this:

n=54 E 6 n(n-1)
54*53*52*51*50*49 = 18595558800

Answer 4

6 place holders _ _ _ _ _ _

54 different numbers (1 - 54)

Therefore each block can contain a number different from it's predecessors

1st block = 54 possible numbers
2nd block = 53 possible numbers
3rd block = 52 possible numbers
4th block = 51 possible numbers
5th block = 50 possible numbers
6th block = 49 possible numbers

multiply them together = 54x53x53x52x50x49 = 18595558800 combinations


PS. Shermin81 --> where did you get the 1 in "54 + 1"?

Answer 5

What you ar looking for is a concept called combinations, just as the problem says.

It is 54! divided by 6! and (54-6)!

To find the answer use Google:

25 827 165

Answer 6

Your base rules were:
1. Combinations must be 6 single digits
2. All six digits must be different
3. Numbers must be from 1-54

Derived constraints:
4. 1-54 basically gives you the single digits 1,2,3,4,5,6,7,8 and 9

So the answer is:
__ __ __ __ __ __
9 x 8 x 7 x 6 x 5 x 4

ie. 60,480 combinations

Answer 7

54^6 because 6 spots _ _ _ _ _ _ and each could have 1-54
so you multiply the chances together

Answer 8

nPr = (n!)/((n - k)!)

54 + 1 = 55

55P6 = (55!)/((55 - 6)!)
55P6 = (55!)/(49!)
55P6 = 55 * 54 * 53 * 52 * 51 * 50
55P6 = 20872566000

Answer 9

Yes, if you want the number of distinct combinations of r objects taken from n distinct objects (r<=n) then you just use:

nCr = n!/(n-r)! r!

In your case 54C6 = 25,827,165

If the order of the r objects mattters, for example (1,2,3,4,5,6) is considered distinct from (6,5,4,3,2,1) then use the permutations formula:

nPr = n!/(n-r)!

Answer 10

18,595,558,800

Sounds like a lottery question to me. You have a 1 in 18,595,558,800 of winning if you purchase 1 ticket.