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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...

You may also find the avlue of x by setting 2/(x+1) = 1/5 and solve x from there. 

Answer was incorrect in the app and I also don't understad tying the the two c's together

Accuracy

8 letters,  3Csand 2As        8!/(3!*2!) = 3360    As the question said

aCuracy   (I tied 2 Cs together)

How many ways can this be arranged

7 letters, 2As

7!/ (2!) = 2520,  but if the little c is on either side of the C then it will be the same so I need to divide by 2

2520/2=1260

So the mumber of ways no two cs are together is     3360-1260 = 2100 ways

If you think this is incorrect then please promote a discussion.

Or at least say so and give reason  (eg answer is rejected in the app)

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

12!/3!*3!*3!*3!

would that be 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?

The two parts of your sequence

682...2043...3404

are the same length.

  !

maybe proofread your question...

Together, they have 80 stickers.

let x be the amount of sticker that Larry has;

That means that Derek has 7x stickers. To get both Derek and Larry to have an equal amount of stickers, you will have to subtract 3x from 7x which is the amount that Derek has, so then they both have 4x stickers. Since Dereck gave 3x stickers, and he also gave 30 stickers, that would been that x = 10. In the begining, if Larry had x stickers, and Derek had 7x stickers, that would  mean that together they have 8x stickers, and since x = 10, that means that together they have
8(10) = 80 stickers.

Sorry for the long explanation.

yes i remember you

Hi asinus: The question is NOT accurately stated !! It has been reversed. You are absolutely right that there is no solution the way the question is transposed! The question should read:

The middle number of a finite sequence of consecutive integers is 2043. The sum of the numbers AFTER 2043 is twice the sum of those BEFORE 2043. How many integers are in the sequence? Thank you.

This is the question I solved !!! I automatically assumed that the question was reversed!.

Hello Guest from Answer #2 #3!

The two parts of your sequence

682...2043...3404

are the same length.

However, the sum of the first part of the sequence should be twice as long as the second part of the sequence.

  !

asinus: Don't understand the solution below yours? proceed as follows:

Simplify equation 1 as follows:

(n + 2042)/2 (2042 - n + 1) = (2042 - n + 1) (2042 - n + 2)

-1/2 (n - 2043) (n + 2042) = (n - 2044) (n - 2043)

-(3 n^2)/2 + (8175 n)/2 - 2089989 = 0

-3/2 (n - 2043) (n - 682) = 0

Divide both sides by - 3/2

(n - 2043) (n - 682) ==0

n ==2043,  or n == 682 - this is the first term of the first half of the sequence

Do the same with equation 2 and should get:

m ==3,404 - this is the last term of the 2nd half of the sequence.

what triangle are we looing for?

Nevermind, solved it. The answer was 32 :)))

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. 

Nevermind i solved it, the answer is 15 :))

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 \(r^2 = \color{brown}\boxed{100}\)

The first digit will represent \(7^3\). The next will represent \(7^2\), then \(7^1\), and finally \(7^0\), so there are \(\color{brown}\boxed{4}\) digits.

As long as the top part \(\sqrt{x-2} \ge 0\) is positive, the function is defined. 

Can you take it from here?

(n + 2042)/2 (2042 - n + 1) = (2042 - n + 1) (2042 - n + 2),.........(1) solve for n
n==682 - this is the first term
 
(m + 2044) / 2 * (m - 2044 +1)==2(m - 2044 +1)(m - 2044 +2),.........(2) solve for m
m ==3,404 - this is the last term

m - n + 1==3,404 - 682 + 1==2,723 terms.

_ _ _ _ _ _ _ _     8 digits

"Exactly five 1's and three 2's"

So, let's just write any 8 digits that work:

1 1 1 1 1 1 2 2 2   

Now, we want all the possible permutations without repetition. 

I.e. we want: 1 1 1 1 1 2 1 2 2 and so on.

How many of these?
It is the same idea as, how many words can be made from the word "ACCURACY"

We have 8 digits, 5 of them are 1's (same thing) and 3 of them are 2's (same thing)
So:     \(\frac{8!}{5!*2!}\)  =168 .

(this is saying, arrange the 8 digits as you want, then we divide by 5! as the 1's are repeated, and divide by 2! as the 2's are repeated.) 

I hope this helps! 

Hi Jason!

Let's write the word "ACCURACY" as below:

A C   U R      Y               (This is just listening the letters, and any repeated letter below each other; thus making counting repeats easier.)
A C  

   C

Total number of letters = 8     (You used 7, not 8) 

Repeated C = 3 

Repeated A = 2 

Therefore ,\(\frac{8!}{3!*2!}\)  = 3360 ways

(We divide the total by the number of the repeated, and accounting for their permutations by using !).

I hope this helps :)!