MHB Is 106 = 9Cx correct for this combination problem?

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The discussion centers on the equation 106 = 9Cx, where participants analyze the validity of this combination problem. It is established that the combination formula is 9!/(x!(9 - x)!), and the correct setup requires appropriate parentheses. Participants conclude that there is no natural number solution to the equation, as none of the combinations equal 106. The original source of the problem is unclear, originating from a Facebook math group, which raises concerns about its accuracy. Ultimately, the consensus is that without the original problem context, resolving the issue is challenging.
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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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