Is 106 = 9Cx correct for this combination problem?

  • Context: MHB 
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Discussion Overview

The discussion revolves around the equation 106 = 9Cx, which is related to combinations in combinatorial mathematics. Participants explore the validity of this equation and its implications, questioning the setup and potential solutions.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation 106 = 9!/x!(9 - x)! and seeks confirmation on its correctness.
  • Another participant questions whether 106 is the correct value for the combination, providing a list of values for 9Cx for x = 1 to 9, none of which equal 106.
  • Several participants discuss the expression 6C2 + 7C4 + 8C5 = 9Cx, with some asserting that this leads to the conclusion that 106 = 9Cx.
  • There is a repeated emphasis on the need for appropriate parentheses in the combination formula, with participants clarifying the correct expression.
  • One participant mentions that there are no natural number solutions to the problem, referencing a tool that provided numeric approximations instead.
  • Another participant expresses skepticism about the accuracy of problems found in social media math groups.

Areas of Agreement / Disagreement

Participants generally disagree on the validity of the equation 106 = 9Cx, with some asserting it cannot be true while others attempt to justify it through calculations. The discussion remains unresolved regarding the existence of a solution.

Contextual Notes

There are limitations in the discussion, including missing assumptions about the context of the problem and the lack of a clear source for the original question, which complicates the evaluation of the claims made.

mathdad
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I found an interesting question online. It involves combination.

106 = 9Cx

I know the combination formula but can someone set it up for me?

Here is my endeavor:

106 = 9!/x!(9 - x)!

Is this correct?
 
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RTCNTC said:
106 = 9Cx
You sure that should be 106?

(9)C(x), x = 1 to 9:
9,36,84,126,126,84,36,9,1
 
Last edited:
6C2 + 7C4 + 8C5 = 9Cx
 
RTCNTC said:
6C2 + 7C4 + 8C5 = 9Cx
Says who?
 
Wilmer said:
Says who?

I found this question online not from a textbook.
 
I gave you the possible solutions: none equals 106.

Can you give us the url of the site?
 
RTCNTC said:
Here is my endeavor:
106 = 9!/x!(9 - x)!
The right side should be 9! / (x!(9 - x))
Do you see why the extra set of brackets is required?

Once more: there is NO SOLUTION = 106
 
RTCNTC said:
106 = 9!/x!(9 - x)!

You need appropriate parentheses. The above should be

9!/(x!(9 - x)!)
 
6C2 + 7C4 + 8C5 = 9Cx

6C2 = 15

7C4 = 35

8C5 = 56

15 + 35 + 56 = 9Cx

50 + 56 = 9Cx

106 = 9Cx

Do you see where 106 comes from?
 
  • #10
RTCNTC said:
6C2 + 7C4 + 8C5 = 9Cx
6C2 = 15
7C4 = 35
8C5 = 56
15 + 35 + 56 = 9Cx
50 + 56 = 9Cx
106 = 9Cx
Do you see where 106 comes from?
Doesn't matter where it comes from.
Look buddy, give us the link to where you found
this problem...else you're on your own...
 
  • #11
I found this problem in a facebook math group but there are so many that I truly don't recall the name of the group. This question is not so important. Thanks anyway.
 
  • #12
Like Wilmer, I found that there is no natural number solution to that problem. W|A was able to return 2 numeric approximations, but with a problem like this, we expect natural numbers. Seeing the original problem would have helped to clear up the issue. :)
 
  • #13
Sorry if I sounded exasperated Mr.RTCNTC.

I've seen quite a few "Facebook math posts" that are inaccurate...
 
  • #14
Can we just move on?
 

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