Solving Combination Problem: 11Cn=330

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SUMMARY

The discussion focuses on solving the combination problem represented by the equation 11Cn = 330. The solution involves using the combination formula nCr = n! / (r!(n-r)!) to derive the equation n!(11-n)! = 133056. Participants concluded that the values of n that satisfy this equation are 4 and 7, achieved through a combination of algebraic manipulation and guess-and-check methods. The discussion emphasizes the importance of simplifying factorials to identify patterns in the problem.

PREREQUISITES
  • Understanding of combination formulas, specifically nCr = n! / (r!(n-r)!)
  • Knowledge of factorial notation and operations
  • Basic algebraic manipulation skills
  • Experience with problem-solving techniques, including guess and check
NEXT STEPS
  • Study the properties of combinations and permutations in combinatorial mathematics
  • Learn advanced techniques for simplifying factorial expressions
  • Explore algebraic methods for solving equations involving factorials
  • Practice additional combination problems to reinforce understanding of nCr
USEFUL FOR

Students studying combinatorial mathematics, educators teaching algebraic concepts, and anyone interested in solving factorial-related problems.

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Homework Statement


Solve for n.
[tex]11_{C}_n=330[/tex]


Homework Equations


[tex]n_{C}_r=\frac{n!}{(n-r)!r!}[/tex]
Sorry if the combination formula looks bad. I don't know how to write the comb. formula with Latex.


The Attempt at a Solution


I solved for n as far as [tex]n!(11-n)!=133056[/tex] How do I go further with this? The answers are 4 or 7.
 
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Honestly the easiest way to do this would be a combination of algebra and guess and check. You should see that both n and 11-n would solve the equation, so finding one answer would give you the answer.

The usual way to reduce factorials is to try to keep as much of the factorial together as possible, which often let's you see patterns easier. For instance:

[tex]_{11}C_n = \frac{11!}{n!(11-n)!} = 330 \implies n!(11-n)! = \frac{11!}{2 \cdot 3 \cdot 5 \cdot 11}[/tex]

You can cancel out the 11 and the 10 on top, leaving you with:

[tex]n!(11-n)! = 3 \cdot 8![/tex]

Now you start with the guess and check. You know that n has to be less than 8.
 

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