BuilderBoi

The only constraint is that the leading digit must be a 3 or greater. 

Can you take it from here?

Step 1: Find the number of ways. Treat this as the "ACCURACY" problem, only we have colors instead of letters. 

Step 2: Find the number of ways there will be 2 green cubes at the bottom. If there are 2 green cubes at the bottom, there are 8 cubes left to order, which consist of 1 red, 2 yellow, 1 green, and 4 blue. Do the same thing as the "ACCURACY" problem. 

Step 3: Subtract the result you got from Step 2 from the result you got from Step 1.

Here's my attempt: 

There are 2 cases to consider: exactly 2 C's are next to each other, or 3 C's are next to each other. 

If 3 C's are next to each other, there are 6 ways to put them. Then, for the remaining 5 digits, there are \({5! \over 2!} = 60\) ways to order them. So, there are \(6 \times 60 = 360\) ways to order this. 

If 2 C's are next to each other, then there are 2 subcases: The 2 C's are on the sides (1 and 2 or 7 and 8) or they are in the center.

If the C's are on the sides, there are 2 ways to choose to put the C's. Then, the third C has 5 choices (can't be right next to the other C's). Then, there are \({5! \over 2!} = 60\) ways to order the remaining digits. So, there are \(2 \times 5 \times 60 = 600\) ways for this case. 

Now, if the C's are in the center, there are 5 ways to put the C's. Then, the third C has 4 choices (can't be on either side). Then, there are \({5! \over 2!} = 60\) ways for the remaining digits. So, there are \(5 \times 4 \times 60 = 1200\) ways for this. 

So, there are \(3360 - 300-600-1200 = \color{brown}\boxed{1260}\) ways. I think...

Yeah, that's what it would. Note that I'm not sure if my logic is correct.\(\)

Here's my attempt:

Let the students be A, B, C, and D. 

We basically have the question "How many ways can you arrange 3 A's, 3 B's, 3 C's, and 3 D's to form a 12 letter word?"

Note that the word will basically be the order you call on the students. So AAABBBCCCDDD means you call on A for the first 3 times, then B for the next 3 questions, then C, and finally D. 

Can you take it from here?

maybe proofread your question...

yes i remember you

what triangle are we looing for?

There are \({8! \over 3! \times 2!} = 3360\) ways to arrange the letters as you found out already.

Use complementary counting and count the ways that the 2 A's can be next to each other. 

If the 2 A's are next to each other, there are 7 spots for them. This gives us \({6! \over 3!} = 120\) ways to arrange the other digits (There are 3 C's).

So, there are \(7 \times 120 = 840 \) ways for the 2 A's to be next to each other. 

Thus, there are \(3360 - 840 = \color{brown}\boxed{2520}\) ways in which no 2 A's are next to each other. 

We have the equation \(x^2 + 10x + y^2 - 8y = 59 \)

Add \((10 \div 2)^2 = 5^2 = 25\) to both sides to complete the square on x and \((-8 \div 2)^2 = (-4)^2 = 16\) to both sides to complete the square on y. 

This gives us \(x^2 + 10x + 25 + y^2 - 8y + 16 = 59 + 25 + 16\)

Completing the square on x and y gives us \((x+5)^2 + (y-4)^2 = 100\), meaning \(r^2 = \color{brown}\boxed{100}\)