MHB How to find combination of values which comes out of P&C formulas

AI Thread Summary
The discussion focuses on finding combinations of elements using the formula C(n, r) = n! / (r!(n-r)!), which calculates the number of ways to choose r elements from a set of n distinguishable objects without regard to order. Participants clarify that while the formula can determine the number of subsets, it does not specify which subsets can be formed. Examples are provided to illustrate how subsets can be represented using binary digits, akin to generating a power set. Additionally, suggestions for algorithms to generate combinations are mentioned, with a recommendation to consult "The Art of Computer Programming" by Knuth for deeper insights. The conversation emphasizes the complexity of generating combinations for larger sets and the utility of algorithmic approaches.
rajemessage
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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
 
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rajemessage said:
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 $1$ ?
 
mathmari said:
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 $\displaystyle{C(n,r)=\frac{n!}{r!(n-r)!}}$
 
rajemessage said:
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: $\displaystyle{C(3,3)=\frac{3!}{3!(3-3)!}=1}$
 
mathmari said:
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
 
Last edited:
mathmari said:
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]
 
Last edited:
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