Is there a general formula for the number of groups that can be created from a finite set

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I came upon the formula $(n^2-n)/2$, which gives the number of unique pairs from n elements. Am wondering if there is a general formula to achieve this for higher number of groupings i.e given 6 elements how many 3 member groups can be made.

combinatorics

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asked Jul 24 at 3:57

Eddy Zavala

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  I believe you are looking for combinations and the binomial coefficient
  – WaveX
  Jul 24 at 4:02
  

  
  
  
  

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I came upon the formula $(n^2-n)/2$, which gives the number of unique pairs from n elements. Am wondering if there is a general formula to achieve this for higher number of groupings i.e given 6 elements how many 3 member groups can be made.

combinatorics

share|cite|improve this question

asked Jul 24 at 3:57

Eddy Zavala

415

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I believe you are looking for combinations and the binomial coefficient
  – WaveX
  Jul 24 at 4:02
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I came upon the formula $(n^2-n)/2$, which gives the number of unique pairs from n elements. Am wondering if there is a general formula to achieve this for higher number of groupings i.e given 6 elements how many 3 member groups can be made.

combinatorics

share|cite|improve this question

asked Jul 24 at 3:57

Eddy Zavala

415

I came upon the formula $(n^2-n)/2$, which gives the number of unique pairs from n elements. Am wondering if there is a general formula to achieve this for higher number of groupings i.e given 6 elements how many 3 member groups can be made.

combinatorics

share|cite|improve this question

asked Jul 24 at 3:57

Eddy Zavala

415

share|cite|improve this question

share|cite|improve this question

asked Jul 24 at 3:57

Eddy Zavala

415

asked Jul 24 at 3:57

Eddy Zavala

415

asked Jul 24 at 3:57

Eddy Zavala

415

415

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I believe you are looking for combinations and the binomial coefficient
  – WaveX
  Jul 24 at 4:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  I believe you are looking for combinations and the binomial coefficient
  – WaveX
  Jul 24 at 4:02
  

  
  
  
  

4

4

I believe you are looking for combinations and the binomial coefficient
– WaveX
Jul 24 at 4:02

I believe you are looking for combinations and the binomial coefficient
– WaveX
Jul 24 at 4:02

add a comment | 

1 Answer
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$n choose k =fracn!(n-k)!k!$, known as "$n$ choose $k$", is the number of ways of choosing $k$ elements from $n$.

Coincidentally (or not so coincidentally), and quite famously (at least among mathematicians) it is also the "binomial coefficient"... you may wish to look it up.

Check that when $k=2$ we get your formula.

So, for $n=6$ and $k=3$, we get $6choose 3=frac6!(6-3)!3!=frac6!3!3!=frac6cdot5cdot 43!=20$.

(Btw, it's pretty easy to see why... For instance, in the last example, there are $6$ choices for the first element of the group, then $5$ choices for the second, and $4$ for the third. This gives $6cdot5cdot4=120$. But there are $6$ ways to "permute", or put the $3$ objects in different orders. That's because we have $3$ choices for the first, $2$ for the second, and $1$ for the third. So we divide by $6$, to arrive at $20$...
groups or "combinations".)

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edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

$n choose k =fracn!(n-k)!k!$, known as "$n$ choose $k$", is the number of ways of choosing $k$ elements from $n$.

Coincidentally (or not so coincidentally), and quite famously (at least among mathematicians) it is also the "binomial coefficient"... you may wish to look it up.

Check that when $k=2$ we get your formula.

So, for $n=6$ and $k=3$, we get $6choose 3=frac6!(6-3)!3!=frac6!3!3!=frac6cdot5cdot 43!=20$.

(Btw, it's pretty easy to see why... For instance, in the last example, there are $6$ choices for the first element of the group, then $5$ choices for the second, and $4$ for the third. This gives $6cdot5cdot4=120$. But there are $6$ ways to "permute", or put the $3$ objects in different orders. That's because we have $3$ choices for the first, $2$ for the second, and $1$ for the third. So we divide by $6$, to arrive at $20$...
groups or "combinations".)

share|cite|improve this answer

edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

add a comment | 

up vote
2
down vote

$n choose k =fracn!(n-k)!k!$, known as "$n$ choose $k$", is the number of ways of choosing $k$ elements from $n$.

Coincidentally (or not so coincidentally), and quite famously (at least among mathematicians) it is also the "binomial coefficient"... you may wish to look it up.

Check that when $k=2$ we get your formula.

So, for $n=6$ and $k=3$, we get $6choose 3=frac6!(6-3)!3!=frac6!3!3!=frac6cdot5cdot 43!=20$.

(Btw, it's pretty easy to see why... For instance, in the last example, there are $6$ choices for the first element of the group, then $5$ choices for the second, and $4$ for the third. This gives $6cdot5cdot4=120$. But there are $6$ ways to "permute", or put the $3$ objects in different orders. That's because we have $3$ choices for the first, $2$ for the second, and $1$ for the third. So we divide by $6$, to arrive at $20$...
groups or "combinations".)

share|cite|improve this answer

edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

add a comment | 

up vote
2
down vote

up vote
2
down vote

$n choose k =fracn!(n-k)!k!$, known as "$n$ choose $k$", is the number of ways of choosing $k$ elements from $n$.

Coincidentally (or not so coincidentally), and quite famously (at least among mathematicians) it is also the "binomial coefficient"... you may wish to look it up.

Check that when $k=2$ we get your formula.

So, for $n=6$ and $k=3$, we get $6choose 3=frac6!(6-3)!3!=frac6!3!3!=frac6cdot5cdot 43!=20$.

(Btw, it's pretty easy to see why... For instance, in the last example, there are $6$ choices for the first element of the group, then $5$ choices for the second, and $4$ for the third. This gives $6cdot5cdot4=120$. But there are $6$ ways to "permute", or put the $3$ objects in different orders. That's because we have $3$ choices for the first, $2$ for the second, and $1$ for the third. So we divide by $6$, to arrive at $20$...
groups or "combinations".)

share|cite|improve this answer

edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

$n choose k =fracn!(n-k)!k!$, known as "$n$ choose $k$", is the number of ways of choosing $k$ elements from $n$.

Coincidentally (or not so coincidentally), and quite famously (at least among mathematicians) it is also the "binomial coefficient"... you may wish to look it up.

Check that when $k=2$ we get your formula.

So, for $n=6$ and $k=3$, we get $6choose 3=frac6!(6-3)!3!=frac6!3!3!=frac6cdot5cdot 43!=20$.

(Btw, it's pretty easy to see why... For instance, in the last example, there are $6$ choices for the first element of the group, then $5$ choices for the second, and $4$ for the third. This gives $6cdot5cdot4=120$. But there are $6$ ways to "permute", or put the $3$ objects in different orders. That's because we have $3$ choices for the first, $2$ for the second, and $1$ for the third. So we divide by $6$, to arrive at $20$...
groups or "combinations".)

share|cite|improve this answer

edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 5:35

edited Jul 24 at 5:35

edited Jul 24 at 5:35

answered Jul 24 at 4:14

Chris Custer

5,4082622

answered Jul 24 at 4:14

Chris Custer

5,4082622

answered Jul 24 at 4:14

Chris Custer

5,4082622

5,4082622

add a comment | 

add a comment | 

 

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