Finding the order of elements in combinations using factorial formulas

[rajemessage]

Dear All,

like subsets {abc} can be found out using binary digits like

000
001
010
etc
etc.

how can i find the order of element for any combination which can be calulated useing following formula.
n!/r!(n-r)!

yours sincerley

 

[mathmari]

Dear All,

like subsets {abc} can be found out using binary digits like

000
001
010
etc
etc.

how can i find the order of element for any combination which can be calulated useing following formula.
n!/r!(n-r)!

yours sincerley

The formula $\displaystyle{C(n,r)=\frac{n!}{r!(n-r)!}}$ shows the number of ways a sample of “$r$” elements can be obtained from a larger set of “$n$” distinguishable objects where order does not count and repetitions are not allowed.Do you maybe want to find how many sets of $3$ elements can be created by using the digits $0$ and $1$ ?

 

[rajemessage]

The formula $\displaystyle{C(n,r)=\frac{n!}{r!(n-r)!}}$ shows the number of ways a sample of “$r$” elements can be obtained from a larger set of “$n$” distinguishable objects where order does not count and repetitions are not allowed.Do you maybe want to find how many sets of $3$ elements can be created by using the digits $0$ and $1$ ?

like power set of {abc} can be found out using 3 digits of binary,
like
abc Subset
0 000 { }
1 001 {c}
2 010 {b}
3 011 {b,c}
4 100 {a}
5 101 {a,c}
6 110 {a,b}
7 111 {a,b,c}

in similar fashion i wanted to know the sets of $\displaystyle{C(n,r)=\frac{n!}{r!(n-r)!}}$

 

[mathmari]

like power set of {abc} can be found out using 3 digits of binary,
like
abc Subset
0 000 { }
1 001 {c}
2 010 {b}
3 011 {b,c}
4 100 {a}
5 101 {a,c}
6 110 {a,b}
7 111 {a,b,c}

in similar fashion i wanted to know the sets of $\displaystyle{C(n,r)=\frac{n!}{r!(n-r)!}}$

You cannot find which subsets, only the number of them.

If you want to find the number of all the subsets of the set $\{a,b,c\}$:

The number of subsets with $0$ elements: $\displaystyle{C(3,0)=\frac{3!}{0!(3-0)!}=1}$

The number of subsets with $1$ elements: $\displaystyle{C(3,1)=\frac{3!}{1!(3-1)!}=\frac{3!}{2!}=3}$

The number of subsets with $2$ elements: $\displaystyle{C(3,2)=\frac{3!}{2!(3-2)!}=\frac{3!}{2!}=3}$

The number of subsets with $3$ elements: $\displaystyle{C(3,3)=\frac{3!}{3!(3-3)!}=1}$

 

[rajemessage]

You cannot find which subsets, only the number of them.

If you want to find the number of all the subsets of the set $\{a,b,c\}$:

The number of subsets with $0$ elements: $\displaystyle{C(3,0)=\frac{3!}{0!(3-0)!}=1}$

The number of subsets with $1$ elements: $\displaystyle{C(3,1)=\frac{3!}{1!(3-1)!}=\frac{3!}{2!}=3}$

The number of subsets with $2$ elements: $\displaystyle{C(3,2)=\frac{3!}{2!(3-2)!}=\frac{3!}{2!}=3}$

The number of subsets with $3$ elements: $\displaystyle{C(3,3)=\frac{3!}{3!(3-3)!}=1}$

how can i find the sets from following situation.
i have three numbers,{1 2 3} which will always be in this order {123},
i want to find out number of cases can be made,
.
but 2 can come at frist position that is before 1 or at second position or at
third position that is after 3.
and all are optional any link will be helpfull.
yours sincerly

 

[awkward]

You cannot find which subsets, only the number of them.
[snip]

Actually, there are many methods for generating the subsets. If you google "algorithms for generating combinations" you will find many hits. Here is one:
Algorithm to return all combinations of k elements from n - Stack Overflow

If you really want to read about the subject in depth, see Volume 4 of "The Art of Computer Programming" by Knuth. Section 7.2.1.3 is titled "Generating all combinations".

[edit] P.S. Clearly you don't want to try this by hand for large numbers of objects. Life is too short. [/edit]