Binomial Coefficient - Factorials Part II

Click For Summary

Homework Help Overview

The problem involves dividing twenty-one people into two teams, the Red Team with 10 members and the Blue Team with 11 members. Participants are exploring the calculation of the number of ways to achieve this division using binomial coefficients.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster suggests that the solution may involve the binomial coefficients ##\binom{21}{10}## or ##\binom{21}{11}##, noting their equivalence. Some participants mention using spreadsheet software for calculations, while others provide alternative methods for computing the coefficients without direct calculation of large factorials.

Discussion Status

Participants are actively discussing various methods to approach the problem, including the use of software tools and alternative mathematical expressions. There is no explicit consensus on a single method, but several productive lines of reasoning are being explored.

Contextual Notes

Some participants express uncertainty about using online calculators and spreadsheet software, indicating a potential constraint in their familiarity with these tools. The discussion also touches on the complexity of calculating large factorials directly.

reenmachine
Gold Member
Messages
514
Reaction score
9

Homework Statement



This one is trickier than the problem in my other thread in my opinion.Twenty-one people are to be divided into two teams , The Red Team and the Blue Team.There will be 10 people on Red Team and 11 people on Blue Team.How many ways to do this?

I am not sure how to solve this problem but I have the feeling it is simply ##\binom{21}{10}## or ##\binom{21}{11}## which are the same thing since: ##\frac{21!}{10!(21-10)!} = \frac{21!}{10!11!}## and ##\frac{21!}{11!(21-11)!} = \frac{21!}{11!10!}##.

Now since 21! is a huge number , I decided to use a big number calculator online.Hopefully this is not forbidden.

Homework Equations



##\frac{21!}{11!(21-11)!} = \frac{51 090 942 171 709 440 000}{39916800 \cdot 3628800} = \frac{51 090 942 171 709 440 000}{144 850 083 840 000} = 352 716##thoughts on this one?

thank you!
 
Last edited:
Physics news on Phys.org
Right again.
Spreadsheet software has functions for this sort of thing. In OpenOffice Calc you'd write =combin(21;10).
 
haruspex said:
Right again.
Spreadsheet software has functions for this sort of thing. In OpenOffice Calc you'd write =combin(21;10).

Thank you!

Unfortunately I am not familiar with spreadsheet software or OpenOffice Calc.
 
I think formally it can be done like this: ##\binom{21}{10}\times\binom{11}{11}##
 
reenmachine said:

Homework Statement



This one is trickier than the problem in my other thread in my opinion.Twenty-one people are to be divided into two teams , The Red Team and the Blue Team.There will be 10 people on Red Team and 11 people on Blue Team.How many ways to do this?

I am not sure how to solve this problem but I have the feeling it is simply ##\binom{21}{10}## or ##\binom{21}{11}## which are the same thing since: ##\frac{21!}{10!(21-10)!} = \frac{21!}{10!11!}## and ##\frac{21!}{11!(21-11)!} = \frac{21!}{11!10!}##.

Now since 21! is a huge number , I decided to use a big number calculator online.Hopefully this is not forbidden.

Homework Equations



##\frac{21!}{11!(21-11)!} = \frac{51 090 942 171 709 440 000}{39916800 \cdot 3628800} = \frac{51 090 942 171 709 440 000}{144 850 083 840 000} = 352 716##


thoughts on this one?

thank you!

If you use software that does exact rational calculations, you can do it more easily as follows:
[tex]{n \choose k} = \frac{n(n-1)(n-2) \cdots (n-k+1)}{k (k-1) (k-2) \cdots 1}<br /> = \frac{n}{k} \cdot \frac{n-1}{k-1} \cdot \frac{n-2}{k-2} \cdots \frac{n-k+1}{1} \\<br /> = \prod_{j=0}^{k-1} \frac{n-j}{k-j}[/tex]
So, in your case
[tex]{21 \choose 10} = \frac{21}{10} \cdot \frac{20}{9} \cdot \frac{19}{8} \cdots \frac{12}{1}[/tex]