BuilderBoi

 \(\dfrac{\sqrt[4]{7}}{\sqrt[3]{7}} = 7 \)

\({\sqrt[4]{7} \over \sqrt[3]{7}} = {7^{1 \over 4} \over 7^{1 \over 3}} = 7^{-1 \over 12}\)

\(\color{brown}\boxed{-{1 \over 12}} \)

If they sum to 2, there are 2 stars and 9 bars which makes for \({11 \choose 9} = 55\) solutions. 

If they sum to 1, there is 1 star and 9 bars which makes for \({10 \choose 9} = 10\) solutions. 

If they sum to 0, there are 0 stars and 9 bars which makes for \({9 \choose 9} = 1\) solution. 

So, there are \(55 + 10 + 1 = \color{brown}\boxed{66} \) solutions.

The triangle inequality states that the two legs (smaller numbers) must have a larger sum than the hypotenuse (largest number)

Can you take it from here?

9 chocolates, 7 friends = 9 stars, 6 bars. So, there are \({15 \choose 6} = 5005 \) combinations.

If Dhruv gets 2, there are 5 chocolates for 6 people, which means there are 7 stars and 5 bars, making for \({12 \choose 5} = 792\) cases

So, the probabilty is \({792 \over 5005} = \color{brown}\boxed{72 \over 455}\) 

There are 2 stars and 10 bars which makes for \({12 \choose 10} = \color{brown}\boxed{66}\) combinations.

By stars and bars, there are 5 bars and 18 stars, so there are \({23 \choose 5} = 33,649\)

But, this overcounts for the cases where x is greater than 10. 

If x is 18, there are 0 stars and 4 bars which makes for \({4 \choose 4} = 1 \) case. 

If x is 17, there are \({5 \choose 4} = 5\) cases

If x is 16, there are \({6 \choose 4} = 15\)

Notice a pattern?

This pattern will continue until \({11 \choose 4}\). Summing all these up gives us \(792\) cases.

So, there are \(33649 - 792 = \color{brown}\boxed{32,857} \) cases.

Cross multiply: \(3(3x-1) = 2(4x-4)\)

Simplify: \(9x - 3 = 8x - 8\)

Subtract 8x from both sides: \(x - 3 = -8\)

Add 3 to both sides: \(x = \color{brown}\boxed{-5}\)