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Unfortunately, that's wrong, but nice try!

The total arrangements is 8!, but the total arrangements where there are 2 same letters next to one another is 7! * 3, since it is 7! for each letter, A, L, and S. Then, there are 6! * 3 scenarios where two of the same letters are next to one another, either A and L, A and S, or L and S. Then, there are 5! arrangements where all letters are next to the other same letter. We then subtract everything, giving us 8! - (7!*3) - (6!*3) - 5!, or 22920.

*answer

In case you guys didn't know, he left the community for unknown reasons. Now Melody is the most hearted person in the community.

I got the answer, thanks. It's eighteen.

Unfortunately, 9 is wrong, but thanks for trying!

The prime factorization of 144 is $2^4 \cdot 3^2$ and the prime factorization of 196 is $2^2 \cdot 7^2$. To count the number of factors of each number, we add 1 to each exponent in the prime factorization and then multiply. Thus, 144 has $(4+1)(2+1)=15$ factors and 196 has $(2+1)(2+1)=9$ factors.

To count the number of factors of either 144 or 196 but not both, we need to count the factors of 144 that are not factors of 196, the factors of 196 that are not factors of 144, and the factors that are common to both.

The prime factors of 144 that are not in 196 are $2^2$ and $3^2$. Thus, there are $(2+1)(2+1)-1=8$ factors of 144 that are not factors of 196.

The prime factors of 196 that are not in 144 are $7^2$. Thus, there is $1$ factor of 196 that is not a factor of 144.

Finally, the factors that are common to both 144 and 196 are those that have prime factors of $2^2$ and $7^2$, of which there is $1$.

Therefore, the total number of positive integers that are factors of either 144 or 196 but not both is $8+1=\boxed{9}$.

That should be the anser, but it's not. Strange. Thank so much anyways! :)

If you look on desmos, the answer is (-3,1)

Let a/3=b

Now, we have b+1=3b+1/b+3 ==> 2b+2+1/b=0

Multiplying both sides by b, 2b^2+2b+1=0

Using the quadratic formula, the roots are -0.5 + 0.5i and -0.5 - 0.5i.

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This has been asked several times.

Retirement?

Try this: 

Let's call the two points A and B, and let O be the center of the circle.

Without loss of generality, we can assume that point A is located at the top of the circle. Then, the distance between point A and any other point B on the circle is given by the length of the arc subtending angle AOB.

If we fix point A and randomly choose a point B, the probability that the distance between A and B is at most 1 is the same as the probability that angle AOB is less than or equal to 60 degrees (since the length of an arc subtending an angle of 60 degrees on a unit circle is 1).

So, let's assume that point A is located at (1, 0) and let's consider the region of the circle to the right of A. The probability that point B falls in this region is 1/2. To find the probability that the angle AOB is less than or equal to 60 degrees, we need to find the length of the arc that subtends this angle and divide it by the total circumference of the circle.

The length of the arc subtending angle AOB is just the length of one-sixth of the circumference of the circle (since angle AOB is 60 degrees and there are 6 such angles in a full circle). The circumference of the circle is 2π, so the length of the arc is (1/6) * 2π = π/3. Therefore, the probability that the distance between the two points is at most 1 is:

(1/2) * (π/3) / (2π) = 1 / 12

So the probability that the distance between the two points is at most 1 is 1/12.

P. S. I did it in another way and got: 3 / 4. But was less sure of that method!

Thank you for your time and effort, but the website says that this is incorrect. 

I have come up with the thought that if two points are chosen that are at most a distance of 1 unit apart from each other, the angle that they make with the center point of the circle is at most 60 degrees (equilateral triangle). This thought might be incorrect, but if it is correct it might have something to do with the answer.

Let the number = N

(5N) / [N + 7] = 2, solve for N

N = 14 / 3

See the answer here:  https://web2.0calc.com/questions/counting_34169

The number of students who take Spanish but not Chinese class decreased by 4.

Let S be the number of students who take Spanish, C be the number of students who take Chinese, and N be the number of students who take neither. We know that S+C+N is constant.

Initially, we have S+C+N0​=N.

After 10 students dropped from Chinese class, we have S+C−10+N1​=N.

We also know that N1​=N0​+4.

Substituting the second equation into the first equation, we get:

S+C−10+N0​+4=N0​

S+C−6=N0​

Subtracting the third equation from the first equation, we get:

S+C+N0​−(S+C−6)=N−(N0​+4)

6=4

Therefore, the number of students who take Spanish but not Chinese class decreased by 4.

The probability is:

[4 C 2] * 5^2 = 150 / 6^4 = 25 / 216

Let's think through this step-by-step:

We have a circle of radius 1

Two points are chosen at random on the circle

We want to find the probability that the distance between the two points is at most 1

To find the probability, we need to calculate:

The number of possibilities where the distance is at most 1

The total number of possibilities

Then divide the first by the second

To count the possibilities where the distance is at most 1:

** The two points can be anywhere on the circle, as long as they are at most 1 unit apart. This forms a "slice" of the circle with an central angle of up to pi radians (since the circle has circumference 2pi and radius 1).
** The area of a slice of a circle is (pi * r^2) * (theta / 360) where r is the radius and theta is the measure of the central angle in degrees. Here, r = 1 and theta = 180 degrees (since pi radians = 180 degrees), so the area of the slice is (pi * 1^2) * (180/360) = pi/2 square units.

** Since the points can be anywhere in this slice, the number of possibilities is (area of slice) = pi/2

The total number of possibilities is the area of the entire circle = pi.

Therefore, the probability is (pi/2) / pi = 1/2

\(a=(bh)/2\)

a=((6)(8))/2

a=48/2

a=24 u^2

After 10 students dropped from the Chinese class, we have: (C - 10) + S + (N + 10) = Total number of students.

That is not true. Just because people dropped from the Chinese class (hence the -10) doesn't mean that they weren't in the Spanish class, so you can't do N+10.

Thank you so much for trying, though. I really appriceate your effort!

BTW, I'm totalkoolnezz, just not signed in.

Let's denote the number of students who take the Chinese class as "C", the number of students who take the Spanish class as "S", and the number of students who take neither class as "N".

Initially, we have: C + S + N = Total number of students We don't know the total number of students, but we do know that N = 27.

After 10 students dropped from the Chinese class, we have: (C - 10) + S + (N + 10) = Total number of students We also know that N + 10 = 31, so N = 21.

Substituting N = 21 into the first equation, we get: C + S + 21 = Total number of students

Subtracting this equation from the second equation, we get: (C - 10) + S + 31 = Total number of students

(C + S + 21 = Total number of students)

-10 + 0 + 10 = 0

Therefore, the change did not impact the number of students who take the Spanish class but not the Chinese class (i.e., the difference between S and (S and C)). The answer is 0.

[1/2/3/4/5 ]   /   [6/7/8/9/10] = 7