MHB Solve the Binomial Theorem Puzzle: Find Missing Member

[Alexstrasuz1]

Screenshot by Lightshot
The translation in binom coefficent of 4th and 10th are mathching each other.
Find the member which doesn't have x in it.
I understand all of it but the part where (n up n-3)=(n up 9) I just don't understand how they got 12 here

 

[topsquark]

Screenshot by Lightshot
The translation in binom coefficent of 4th and 10th are mathching each other.
Find the member which doesn't have x in it.
I understand all of it but the part where (n up n-3)=(n up 9) I just don't understand how they got 12 here

I'm not sure what the first question is asking.

For the second we would have to have
[math](x^2)^i \cdot \left ( \frac{1}{x} \right ) ^{n - i} = 1[/math]

Which leads to 2i = n - i thus i = n/3. So this will only happen for n divisible by 3.

For the third you have
[math]{n \choose n - 3} = {n \choose 9}[/math]

We can simply compare the bottom element on both sides and conclude that n - 3 = 9. On the other hand if we use the definition of the binomial coefficient we get the equation
[math](n - 3) \cdot (n - 4) \cdot (n - 5) \cdot (n - 6) \cdot (n - 7) \cdot (n - 8) = 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4[/math]

Again by comparison we can find that n = 12. On the other hand this is a 6th degree polynomial in n and can potentially have at least two real solutions since we know it already has one. I don't know how to analyze this in general, but WA gives another real root as n = -1. The other four (WA missed one) are complex.

-Dan