Determine how many groups of a given number are in a entire set

  • Thread starter Thread starter slink
  • Start date Start date
  • Tags Tags
    Groups Set
Click For Summary
SUMMARY

The discussion focuses on calculating the number of combinations of a given number in a set, specifically using the formula for combinations denoted as C(n, r). The formula is defined as C(n, r) = n! / (r!(n-r)!), where n represents the total number of items in the set and r represents the number of items to choose. An example provided illustrates that C(5, 3) equals 10, confirming that there are 10 ways to choose 3 items from a set of 5. The conversation highlights the importance of understanding combinatorics for solving such problems.

PREREQUISITES
  • Understanding of factorial notation (n!)
  • Basic knowledge of combinatorics
  • Familiarity with mathematical notation for combinations
  • Ability to perform basic arithmetic operations
NEXT STEPS
  • Study the principles of combinatorics in detail
  • Learn how to derive and apply the combinations formula in various scenarios
  • Explore advanced topics such as permutations and their differences from combinations
  • Practice solving problems involving combinations using different values of n and r
USEFUL FOR

This discussion is beneficial for students studying mathematics, educators teaching combinatorics, and anyone interested in solving problems related to combinations and set theory.

slink
Messages
1
Reaction score
0
Hello all. I'm looking for an equation, one that I use to know how to figure out(but alas I am getting old and senile), that will alow me to determine how many groups of a given number are in a entire set.

For example; in a set of 5, how many possible groups of 3 would there be? The answer is 10 sets of 3 in a set of 5.

So, if anyone knows an equation for this, I would be very greatful. TIA.

James
 
Physics news on Phys.org
You're thinking of a combination. If I write the combination of n things taken r at a time like this: C(n,r), the formula is:

[tex]C(n,r) = \frac{n!}{r!(n-r)!}[/tex]

So in your example:
[tex]C(5,3) = \frac{5!}{3!(5-3)!} = \frac{5!}{3!2!} = \frac{5\cdot 4}{2} = 10[/tex]
 
How about this one??[itex]C_{5}^{3}[/itex]

Did u study combinatorics in school??

Daniel.
 

Similar threads

Replies
9
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
Replies
16
Views
3K
  • · Replies 26 ·
Replies
26
Views
2K
Replies
5
Views
4K
  • · Replies 18 ·
Replies
18
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
Replies
9
Views
4K