Solve Combination Problem: C(n,3) = C(n,8)

  • Thread starter DecayProduct
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In summary, the conversation revolved around the equation C(n,3) = C(n,8) and how to solve for the value of n. One approach was to write the equation in terms of expanded factorials and cancel out common terms, leading to a fifth degree equation. Another approach was to recognize that for the equation to hold, a + b must equal n for C(n,a) = C(n,b).
  • #1
DecayProduct
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Homework Statement


C(n,3) = C(n,8)


Homework Equations



n!/(n-3)!3! = n!/(n-8)!8!

The Attempt at a Solution



My attempts at a solution are many and varied. But, I figure my problem is with the algebraic operations on factorials. Logically, since the numerators are both n!, I figure that I can ignore that and just work on (n-3)!3! = (n-8)!8!

I boiled this down to (n-3)!/(n-8)! = 6720. I solved this based on the fact that I know that 8!/3! = 6720. Therefore, (n-3)! = 8!, and n = 11. But this was simply guess work on my part. What I'm confused about is, what is the proper way to solve the problem? How do I perform algebra on these factorials?
 
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  • #2
well, if u want a full on algebraic way then u basically write (n-3)! as

(n-3)(n-4)(n-5)(n-6)(n-7)(n-8)!

now cancel the (n-8)! from the numerator and the denominator...

u get a fifth degree equation and u solve that!
 
  • #3
One way would be to write the factorials in their expanded forms.
Like (n-3)! = (n-3)(n-4)(n-5)(n-6)(n-7)(n-8)!
 
  • #4
DecayProduct said:
C(n,3) = C(n,8)

Hi DecayProduct! :smile:

I'd just write the answer without giving a reason …

C(n,a) = C(n,b) only if a + b = n seems too obvious to require explanation.

(If you want an explanation, I suppose you could easily show that C(n,a) is monotone increasing in a until half-way, and then monotone decreasing)
 

Related to Solve Combination Problem: C(n,3) = C(n,8)

1. What does "C(n,3) = C(n,8)" mean?

This is a combination problem where we are trying to find the number of ways to choose 3 objects from a group of n objects, and the result is equal to the number of ways to choose 8 objects from the same group.

2. How do you solve this combination problem?

This problem can be solved using the formula C(n,r) = n! / (r!(n-r)!), where n is the total number of objects and r is the number of objects being chosen. In this case, we would set n equal to the number of objects in the group and solve for r by setting C(n,3) = C(n,8).

3. Can you provide an example of solving this type of combination problem?

Sure, let's say we have a group of 10 people and we want to choose 3 people to form a committee. The number of ways to do this would be C(10,3) = 10! / (3!(10-3)!) = 120. Now, let's say we want to choose 8 people to form a different committee from the same group of 10 people. The number of ways to do this would be C(10,8) = 10! / (8!(10-8)!) = 45. As we can see, C(10,3) = C(10,8) = 120 = 45.

4. Are there any other ways to solve this combination problem?

Yes, another way to solve this problem is by using the binomial coefficient formula, which is n! / (r!(n-r)!). In this case, we would set n equal to the number of objects in the group and solve for r by setting C(n,3) = C(n,8).

5. What are some real-life applications of this type of combination problem?

Combination problems can be used in various fields, such as mathematics, engineering, finance, and computer science. Some real-life applications include determining the number of possible combinations for a password, calculating the probability of certain events occurring, and solving problems in statistics and probability.

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