Possible combination calculation

[malraff]

hi
is there an equation i can use to work out combination possibilities from say 4 numbers?
i am not interested in unique in order eg
1234 1243 4321 etc

but rather

1234
123
124
234
12
13
14
1
any ideas?

mal

 

[Deleted member 4993]

hi
is there an equation i can use to work out combination possibilities from say 4 numbers?
i am not interested in unique in order eg
1234 1243 4321 etc

but rather

1234
123
124
234
12
13
14
1
any ideas?

mal

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this assignment.

For your calculation, is 14 same as 41?

Is 112 allowed as a possible number?

 

[malraff]

hi
i am trying to replicate what the following site does

Combinations and Permutations Calculator

www.mathsisfun.com

where order is not important and repetition is not allowed
re your question, yes 1,4 is the same as 4,1
112 is not allowed

obviously with 4 numbers its easy to do manually and get the 15 results possible
{1,2,3,4}
{1,2,3} {1,2,4} {1,3,4} {2,3,4}
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4}
{1} {2} {3} {4}
, but id like to model with 9 or 10 digits which goes into hundreds/thousands of results

 

[Deleted member 4993]

hi
i am trying to replicate what the following site does

Combinations and Permutations Calculator

www.mathsisfun.com

where order is not important and repetition is not allowed
re your question, yes 1,4 is the same as 4,1
112 is not allowed

obviously with 4 numbers its easy to do manually and get the 15 results possible
{1,2,3,4}
{1,2,3} {1,2,4} {1,3,4} {2,3,4}
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4}
{1} {2} {3} {4}
, but id like to model with 9 or 10 digits which goes into hundreds/thousands of results

You should try to derive the formula yourself. From your expected answers for 4 numbers:

{1,2,3,4}................................................................. # of choice choosing 4 out 4........................................= 1
{1,2,3} {1,2,4} {1,3,4} {2,3,4}............................ # of choice choosing 3 out 4.......................................= 4
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4}...................... # of choice choosing 2 out 4...................................... = 6
{1} {2} {3} {4}........................................................ # of choice choosing 1 out 4........................................= 4

Does that ring a bell?

If not, then try the same exercise with 3 numbers and 5 numbers.

Does that ring a bell?

continue.....

 

[pka]

obviously with 4 numbers its easy to do manually and get the 15 results possible
{1,2,3,4}
{1,2,3} {1,2,4} {1,3,4} {2,3,4}
{1,2} {1,3} {1,4} {2,3} {2,4} {3,4}
{1} {2} {3} {4}
, but id like to model with 9 or 10 digits which goes into hundreds/thousands of results

What you have listed above are the nonempty subsets of the set \(\{1,22,3,4\}\).
If you list the same collection from the set of ten digits, \(\{0,1,2,3,4,5,6,7,8,9\}\) there are \(2^{10}-1=1023\) such subsets.
Given any set of \(n\) elements there are \(2^n-1\) nonempty subsets of the set.

 

[malraff]

ah so simple but so good thanks so much for that pka, i really appreciate it