Combinations Formula .... what is it again ?

[deftron]

I need the one where objects can be repteated..


For example, 10 objects into a group of 3 (where they can be repeated)

i.e. numbers 0 through 9 ... groups 000 - 999

Obviously it's 1000 available combinations, buts whats the formula for figuring it out?

 

[bonkers325]

nm

 

[deftron]

Originally posted by: bonkers325
10 choose 3

=

10! / (3!7!)

which is 120 different combinations.

thats non repeating formula , tho

 

[Fenixgoon]

10 objects, using all 10 including repeats = 10! (10x9x8x7....x1)

10 objects, using 9/8/7/6/5/X = the first X numbers of 10!

10 objects, using X, no repeats = 10!/X! (i think)

cant remember no repeats using all objects

 

[deftron]

None of those work....


Basically prove there are 1000 combinations in a Pick 3 lottery

 

[Soccer55]

Originally posted by: Fenixgoon
10 objects, using all 10 including repeats = 10! (10x9x8x7....x1)

Wouldn't this one be 10^10? Cause you have 10 objects for each of the 10 positions.

Using all 10 objects, no repeats would be 10!

-Tom

 

[statik213]

Originally posted by: deftron
I need the one where objects can be repteated..


For example, 10 objects into a group of 3 (where they can be repeated)

i.e. numbers 0 through 9 ... groups 000 - 999

Obviously it's 1000 available combinations, buts whats the formula for figuring it out?

x repeating unique objects into y groups => x^y,

 

[deftron]

Originally posted by: statik213

Originally posted by: deftron
I need the one where objects can be repteated..


For example, 10 objects into a group of 3 (where they can be repeated)

i.e. numbers 0 through 9 ... groups 000 - 999

Obviously it's 1000 available combinations, buts whats the formula for figuring it out?

x repeating unique objects into y groups => x^y,

Thanks ... knew it was easy.. but forgot 😱

 

[bonkers325]

since it is repeatable, its basically 10 choose 1 three times.

so [10!/ (1!9!)]^3

or maybe im stuck in combination mode 😛

 

[Fenixgoon]

Originally posted by: Soccer55

Originally posted by: Fenixgoon
10 objects, using all 10 including repeats = 10! (10x9x8x7....x1)

Wouldn't this one be 10^10? Cause you have 10 objects for each of the 10 positions.

Using all 10 objects, no repeats would be 10!

-Tom

! = factorial, multipling every number down to one...

so 10! = 10x9x8x7x6x5x4x3x2x1, which would give you all possible combinations for 10 objects, 10 used, and including repeats

 

[Soccer55]

Originally posted by: Fenixgoon

Originally posted by: Soccer55

Originally posted by: Fenixgoon
10 objects, using all 10 including repeats = 10! (10x9x8x7....x1)

Wouldn't this one be 10^10? Cause you have 10 objects for each of the 10 positions.

Using all 10 objects, no repeats would be 10!

-Tom

! = factorial, multipling every number down to one...

so 10! = 10x9x8x7x6x5x4x3x2x1, which would give you all possible combinations for 10 objects, 10 used, and including repeats

How so? With your method of factorial, you're saying that you have 10 choices for the first slot, 9 choices for the second, etc. By virtue of not having 10 choices in the second slot, aren't you eliminating the possibility for one of the 10 objects from occupying that spot?

-Tom

 

[Fenixgoon]

Originally posted by: Soccer55

Originally posted by: Fenixgoon

Originally posted by: Soccer55

Originally posted by: Fenixgoon
10 objects, using all 10 including repeats = 10! (10x9x8x7....x1)

Wouldn't this one be 10^10? Cause you have 10 objects for each of the 10 positions.

Using all 10 objects, no repeats would be 10!

-Tom

! = factorial, multipling every number down to one...

so 10! = 10x9x8x7x6x5x4x3x2x1, which would give you all possible combinations for 10 objects, 10 used, and including repeats

How so? With your method of factorial, you're saying that you have 10 choices for the first slot, 9 choices for the second, etc. By virtue of not having 10 choices in the second slot, aren't you eliminating the possibility for one of the 10 objects from occupying that spot?

-Tom

you can have 10 10's, 9 10's and then 1-9, 8 10's then 2 numbers, etc.

 

[DrPizza]

More simply, instead of a combinations formula, it's merely the counting principle...

10 ways to choose the first digit times 10 ways to choose the 2nd digit times 10 ways to choose the third digit.

 

[upsciLLion]

buttz + peenerz = qtf