💡Okay, so I have this equation to solve: [(x - 2)^4 + (x - 6)^4 = -16.] Hmm, that looks a bit complicated because of the fourth powers. I remember that solving equations with higher powers can sometimes be tricky, especially when they're set equal to a negative number. Since the right side is negative, I suspect that the solutions might be complex numbers because even powers of real numbers are always non-negative. So, maybe I need to work with complex numbers here.First, I notice that the equation has two terms: ((x - 2)^4) and ((x - 6)^4). The numbers 2 and 6 are both on the real number line, and they are 4 units apart. Maybe I can simplify this by making a substitution. Let me think, if I let (z = x - 4), then 4 is the midpoint between 2 and 6. That might help because it centers the equation around zero, which could make the algebra simpler.So, if (z = x - 4), then (x = z + 4). Let me substitute this into the equation:[(z + 4 - 2)^4 + (z + 4 - 6)^4 = -16.]Simplifying inside the parentheses:[(z + 2)^4 + (z - 2)^4 = -16.]Okay, that looks a bit better. Now, I have two terms with (z) shifted by 2 and -2. Maybe I can expand these terms to see if they can be combined or simplified further.Let me recall how to expand ((a + b)^4). The binomial expansion formula is:[(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.]So, applying this to both ((z + 2)^4) and ((z - 2)^4):First, expand ((z + 2)^4):[(z + 2)^4 = z^4 + 4z^3(2) + 6z^2(2)^2 + 4z(2)^3 + (2)^4][= z^4 + 8z^3 + 24z^2 + 32z + 16.]Next, expand ((z - 2)^4):[(z - 2)^4 = z^4 + 4z^3(-2) + 6z^2(-2)^2 + 4z(-2)^3 + (-2)^4][= z^4 - 8z^3 + 24z^2 - 32z + 16.]Now, add these two expanded forms together:[(z^4 + 8z^3 + 24z^2 + 32z + 16) + (z^4 - 8z^3 + 24z^2 - 32z + 16)]Let me combine like terms:- (z^4 + z^4 = 2z^4)- (8z^3 - 8z^3 = 0)- (24z^2 + 24z^2 = 48z^2)- (32z - 32z = 0)- (16 + 16 = 32)So, adding them all up:[2z^4 + 48z^2 + 32 = -16.]Hmm, okay. Let me write that as:[2z^4 + 48z^2 + 32 = -16.]To simplify this equation, I can subtract -16 from both sides to bring all terms to one side:[2z^4 + 48z^2 + 32 + 16 = 0.][2z^4 + 48z^2 + 48 = 0.]Wait, that doesn't seem right. Let me check my arithmetic. The original equation after substitution was:[(z + 2)^4 + (z - 2)^4 = -16.]After expanding and adding, I had:[2z^4 + 48z^2 + 32 = -16.]So, subtracting -16 from both sides should be:[2z^4 + 48z^2 + 32 + 16 = 0][2z^4 + 48z^2 + 48 = 0.]Wait, that's correct. So, I can factor out a 2 from all terms:[2(z^4 + 24z^2 + 24) = 0.]Divide both sides by 2:[z^4 + 24z^2 + 24 = 0.]Hmm, this is a quartic equation, but it's quadratic in terms of (z^2). Let me set (w = z^2), so the equation becomes:[w^2 + 24w + 24 = 0.]Now, I can solve this quadratic equation for (w). Using the quadratic formula:[w = frac{-24 pm sqrt{(24)^2 - 4 cdot 1 cdot 24}}{2 cdot 1}][= frac{-24 pm sqrt{576 - 96}}{2}][= frac{-24 pm sqrt{480}}{2}][= frac{-24 pm sqrt{16 cdot 30}}{2}][= frac{-24 pm 4sqrt{30}}{2}][= -12 pm 2sqrt{30}.]Wait, that doesn't seem right because I was expecting complex solutions, but this gives me real numbers. Did I make a mistake somewhere?Let me go back. The original equation after substitution was:[(z + 2)^4 + (z - 2)^4 = -16.]When I expanded and added, I got:[2z^4 + 48z^2 + 32 = -16.]Subtracting -16:[2z^4 + 48z^2 + 48 = 0.]Divide by 2:[z^4 + 24z^2 + 24 = 0.]Let me double-check the expansion of ((z + 2)^4) and ((z - 2)^4). Maybe I made a mistake there.Expanding ((z + 2)^4):First, ((z + 2)^2 = z^2 + 4z + 4).Then, ((z^2 + 4z + 4)^2 = z^4 + 8z^3 + 24z^2 + 32z + 16). Wait, is that correct?Wait, ((z^2 + 4z + 4)^2 = z^4 + 8z^3 + (16z^2 + 8z^2) + ... Hmm, maybe I miscalculated the coefficients.Let me do it step by step:[(z^2 + 4z + 4)(z^2 + 4z + 4)][= z^2(z^2 + 4z + 4) + 4z(z^2 + 4z + 4) + 4(z^2 + 4z + 4)][= z^4 + 4z^3 + 4z^2 + 4z^3 + 16z^2 + 16z + 4z^2 + 16z + 16][= z^4 + (4z^3 + 4z^3) + (4z^2 + 16z^2 + 4z^2) + (16z + 16z) + 16][= z^4 + 8z^3 + 24z^2 + 32z + 16.]Okay, that seems correct.Similarly, ((z - 2)^4 = z^4 - 8z^3 + 24z^2 - 32z + 16).Adding them together:[(z^4 + 8z^3 + 24z^2 + 32z + 16) + (z^4 - 8z^3 + 24z^2 - 32z + 16)][= 2z^4 + 48z^2 + 32.]Yes, that's correct. So, the equation is:[2z^4 + 48z^2 + 32 = -16.]Subtracting -16:[2z^4 + 48z^2 + 48 = 0.]Divide by 2:[z^4 + 24z^2 + 24 = 0.]So, setting (w = z^2), we get:[w^2 + 24w + 24 = 0.]Using the quadratic formula:[w = frac{-24 pm sqrt{24^2 - 4 cdot 1 cdot 24}}{2}][= frac{-24 pm sqrt{576 - 96}}{2}][= frac{-24 pm sqrt{480}}{2}][= frac{-24 pm 4sqrt{30}}{2}][= -12 pm 2sqrt{30}.]Wait, but (sqrt{30}) is a real number, so (w) is real. But (w = z^2), and (z) is a complex number because the original equation equals -16, which is negative. So, (z^2) must be negative, meaning (w) should be negative. But (-12 + 2sqrt{30}) is approximately (-12 + 10.954 = -1.046), which is negative, and (-12 - 2sqrt{30}) is approximately (-12 - 10.954 = -22.954), which is also negative. So, both solutions for (w) are negative, which is fine because (z^2) can be negative in complex numbers.So, (z^2 = -12 + 2sqrt{30}) or (z^2 = -12 - 2sqrt{30}).Wait, but (-12 + 2sqrt{30}) is approximately (-12 + 10.954 = -1.046), which is negative, so (z^2) is negative, which is okay because we're dealing with complex numbers.So, (z = pm sqrt{-12 + 2sqrt{30}}) or (z = pm sqrt{-12 - 2sqrt{30}}).But (sqrt{-12 + 2sqrt{30}}) can be written as (isqrt{12 - 2sqrt{30}}), and similarly for the other term.Wait, let me think. If (w = z^2 = -12 + 2sqrt{30}), then (z = pm sqrt{w} = pm sqrt{-12 + 2sqrt{30}}). But (-12 + 2sqrt{30}) is negative because (2sqrt{30} approx 10.954), so (-12 + 10.954 approx -1.046), which is negative. So, (sqrt{-1.046}) is imaginary.Similarly, (-12 - 2sqrt{30}) is more negative, so its square root is also imaginary.So, let me write (z) as:(z = pm sqrt{-12 + 2sqrt{30}} = pm isqrt{12 - 2sqrt{30}})and(z = pm sqrt{-12 - 2sqrt{30}} = pm isqrt{12 + 2sqrt{30}}).Wait, but (sqrt{12 - 2sqrt{30}}) is a real number because (12 - 2sqrt{30} approx 12 - 10.954 = 1.046), which is positive. Similarly, (12 + 2sqrt{30} approx 22.954), which is positive.So, (z = pm isqrt{12 - 2sqrt{30}}) and (z = pm isqrt{12 + 2sqrt{30}}).But wait, that seems a bit messy. Maybe I can simplify (sqrt{12 pm 2sqrt{30}}).Let me see if (sqrt{12 pm 2sqrt{30}}) can be expressed in the form (sqrt{a} pm sqrt{b}).Assume that (sqrt{12 + 2sqrt{30}} = sqrt{a} + sqrt{b}). Then, squaring both sides:(12 + 2sqrt{30} = a + b + 2sqrt{ab}).So, we have:(a + b = 12)and(2sqrt{ab} = 2sqrt{30}), which implies (sqrt{ab} = sqrt{30}), so (ab = 30).We need two numbers (a) and (b) such that (a + b = 12) and (ab = 30).Let me solve for (a) and (b). The quadratic equation would be (x^2 - 12x + 30 = 0).Using the quadratic formula:(x = frac{12 pm sqrt{144 - 120}}{2} = frac{12 pm sqrt{24}}{2} = frac{12 pm 2sqrt{6}}{2} = 6 pm sqrt{6}).So, (a = 6 + sqrt{6}) and (b = 6 - sqrt{6}).Therefore,(sqrt{12 + 2sqrt{30}} = sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}}).Wait, that doesn't seem right because when I square (sqrt{a} + sqrt{b}), I get (a + b + 2sqrt{ab}), which should equal (12 + 2sqrt{30}). But if (a = 6 + sqrt{6}) and (b = 6 - sqrt{6}), then (a + b = 12) and (ab = (6)^2 - (sqrt{6})^2 = 36 - 6 = 30), which is correct. So, yes, (sqrt{12 + 2sqrt{30}} = sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}}).Similarly, for (sqrt{12 - 2sqrt{30}}), let's assume it's (sqrt{a} - sqrt{b}).Squaring both sides:(12 - 2sqrt{30} = a + b - 2sqrt{ab}).So, (a + b = 12) and (2sqrt{ab} = 2sqrt{30}), so (sqrt{ab} = sqrt{30}), (ab = 30).Same as before, so (a = 6 + sqrt{6}), (b = 6 - sqrt{6}).Thus,(sqrt{12 - 2sqrt{30}} = sqrt{6 + sqrt{6}} - sqrt{6 - sqrt{6}}).But this seems complicated, and I'm not sure if it's necessary. Maybe it's better to just leave it as (sqrt{12 pm 2sqrt{30}}).Alternatively, perhaps I made a mistake earlier in the substitution or expansion. Let me double-check.Wait, when I set (z = x - 4), then (x = z + 4). Substituting into ((x - 2)^4 + (x - 6)^4), we get ((z + 2)^4 + (z - 2)^4). That seems correct.Expanding both, I got:[(z + 2)^4 = z^4 + 8z^3 + 24z^2 + 32z + 16][(z - 2)^4 = z^4 - 8z^3 + 24z^2 - 32z + 16]Adding them together:[2z^4 + 48z^2 + 32]Which equals -16, so:[2z^4 + 48z^2 + 32 = -16][2z^4 + 48z^2 + 48 = 0][z^4 + 24z^2 + 24 = 0]Yes, that's correct.So, (w = z^2), equation becomes (w^2 + 24w + 24 = 0), solutions are (w = -12 pm 2sqrt{30}).So, (z = pm sqrt{-12 pm 2sqrt{30}}).But perhaps I can factor the quartic equation differently. Let me try to factor (z^4 + 24z^2 + 24).Looking for factors of the form ((z^2 + a)(z^2 + b)). Then, (a + b = 24) and (ab = 24).Looking for two numbers that add up to 24 and multiply to 24. Hmm, 2 and 22: 2+22=24, 2*22=44. Not 24. 3 and 21: 3+21=24, 3*21=63. 4 and 20: 4+20=24, 4*20=80. 6 and 18: 6+18=24, 6*18=108. 8 and 16: 8+16=24, 8*16=128. 12 and 12: 12+12=24, 12*12=144. None of these give 24. So, it doesn't factor nicely with integer coefficients. So, maybe I need to use the quadratic formula as I did before.So, (z^2 = -12 pm 2sqrt{30}), so (z = pm sqrt{-12 pm 2sqrt{30}}).But perhaps I can write this as (z = pm sqrt{ - (12 mp 2sqrt{30}) }), which is (z = pm i sqrt{12 mp 2sqrt{30}}).So, (z = pm i sqrt{12 - 2sqrt{30}}) and (z = pm i sqrt{12 + 2sqrt{30}}).Now, let me compute (sqrt{12 - 2sqrt{30}}) and (sqrt{12 + 2sqrt{30}}).Wait, earlier I tried to express (sqrt{12 pm 2sqrt{30}}) as (sqrt{a} pm sqrt{b}), and found that (a = 6 + sqrt{6}) and (b = 6 - sqrt{6}). So,(sqrt{12 + 2sqrt{30}} = sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}})and(sqrt{12 - 2sqrt{30}} = sqrt{6 + sqrt{6}} - sqrt{6 - sqrt{6}}).But this seems more complicated than helpful. Maybe it's better to just leave it as is.So, (z = pm i sqrt{12 - 2sqrt{30}}) and (z = pm i sqrt{12 + 2sqrt{30}}).Now, recalling that (z = x - 4), so (x = z + 4).Thus, the solutions for (x) are:(x = 4 pm i sqrt{12 - 2sqrt{30}}) and (x = 4 pm i sqrt{12 + 2sqrt{30}}).But wait, earlier I thought the solutions were (4 pm 2isqrt{3}) and (4 pm isqrt{2}). That seems different from what I'm getting now. Did I make a mistake in the substitution or expansion?Wait, let me check my initial substitution again. I set (z = x - 4), so (x = z + 4). Then, ((x - 2) = (z + 2)) and ((x - 6) = (z - 2)). That's correct.Expanding ((z + 2)^4 + (z - 2)^4), I got (2z^4 + 48z^2 + 32). That seems correct.Setting equal to -16, so (2z^4 + 48z^2 + 32 = -16), which simplifies to (2z^4 + 48z^2 + 48 = 0), then dividing by 2: (z^4 + 24z^2 + 24 = 0). That's correct.So, the quartic equation is correct, and solving it gives (z^2 = -12 pm 2sqrt{30}), so (z = pm isqrt{12 mp 2sqrt{30}}).Wait, but in the initial problem, the user's solution was different. They had (z^4 + 24z^2 + 24 = 0) factoring as ((z^2 + 12)(z^2 + 2) = 0), which would give (z^2 = -12) or (z^2 = -2), leading to (z = pm 2isqrt{3}) or (z = pm isqrt{2}).But according to my calculations, the quartic doesn't factor into ((z^2 + 12)(z^2 + 2)). Let me check that.If I multiply ((z^2 + 12)(z^2 + 2)), I get (z^4 + 14z^2 + 24), which is not the same as (z^4 + 24z^2 + 24). So, that factorization is incorrect. Therefore, the user's initial solution was wrong.So, the correct factorization is not possible with integer coefficients, and we need to solve it using the quadratic formula, leading to (z^2 = -12 pm 2sqrt{30}).Therefore, the solutions for (x) are:(x = 4 pm isqrt{12 - 2sqrt{30}}) and (x = 4 pm isqrt{12 + 2sqrt{30}}).But perhaps I can simplify (sqrt{12 pm 2sqrt{30}}) further.Let me try to express (sqrt{12 + 2sqrt{30}}) as (sqrt{a} + sqrt{b}).Assume (sqrt{12 + 2sqrt{30}} = sqrt{a} + sqrt{b}).Squaring both sides:(12 + 2sqrt{30} = a + b + 2sqrt{ab}).So, (a + b = 12) and (2sqrt{ab} = 2sqrt{30}), which implies (sqrt{ab} = sqrt{30}), so (ab = 30).We need (a + b = 12) and (ab = 30). Solving this system:The quadratic equation is (x^2 - 12x + 30 = 0).Using the quadratic formula:(x = frac{12 pm sqrt{144 - 120}}{2} = frac{12 pm sqrt{24}}{2} = frac{12 pm 2sqrt{6}}{2} = 6 pm sqrt{6}).So, (a = 6 + sqrt{6}) and (b = 6 - sqrt{6}).Therefore,(sqrt{12 + 2sqrt{30}} = sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}}).Similarly, for (sqrt{12 - 2sqrt{30}}), let me assume it's (sqrt{a} - sqrt{b}).Squaring both sides:(12 - 2sqrt{30} = a + b - 2sqrt{ab}).So, (a + b = 12) and (2sqrt{ab} = 2sqrt{30}), which again gives (ab = 30).Thus, (a = 6 + sqrt{6}) and (b = 6 - sqrt{6}).Therefore,(sqrt{12 - 2sqrt{30}} = sqrt{6 + sqrt{6}} - sqrt{6 - sqrt{6}}).So, substituting back into (z):(z = pm i (sqrt{6 + sqrt{6}} - sqrt{6 - sqrt{6}})) and (z = pm i (sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}})).This seems quite complicated, but it's exact. Alternatively, I can leave the solutions in terms of (sqrt{12 pm 2sqrt{30}}).So, the solutions for (x) are:(x = 4 pm isqrt{12 - 2sqrt{30}}) and (x = 4 pm isqrt{12 + 2sqrt{30}}).Alternatively, using the expressions above, I can write:(x = 4 pm i (sqrt{6 + sqrt{6}} - sqrt{6 - sqrt{6}})) and (x = 4 pm i (sqrt{6 + sqrt{6}} + sqrt{6 - sqrt{6}})).But this might not be necessary unless the problem specifically asks for a certain form.Alternatively, perhaps I made a mistake in the substitution. Let me try a different substitution.Let me set (y = x - 4), so (x = y + 4). Then, ((x - 2) = y + 2) and ((x - 6) = y - 2). So, the equation becomes:[(y + 2)^4 + (y - 2)^4 = -16.]Which is the same as before. So, no difference.Alternatively, maybe I can use symmetry. Notice that ((x - 2)^4 + (x - 6)^4) is symmetric around (x = 4), so substituting (y = x - 4) makes sense.Alternatively, perhaps I can let (u = x - 4), so (x = u + 4), and then the equation becomes:[(u + 2)^4 + (u - 2)^4 = -16.]Which is the same as before.Alternatively, perhaps I can consider that ((x - 2)^4 + (x - 6)^4) is always non-negative for real (x), so the equation has no real solutions, only complex ones.But I already considered that.Alternatively, perhaps I can write the equation as:[(x - 2)^4 + (x - 6)^4 = -16.]Let me consider that ((x - 2)^4 = a) and ((x - 6)^4 = b), so (a + b = -16). Since both (a) and (b) are fourth powers, they are non-negative for real (x), but since their sum is negative, both (a) and (b) must be negative, meaning (x - 2) and (x - 6) are complex numbers.So, (x - 2 = c) where (c^4 = a), and (x - 6 = d) where (d^4 = b), and (c + d = -4) because (x - 2 + x - 6 = 2x - 8 = 2(x - 4)), but I'm not sure if that helps.Alternatively, perhaps I can set (c = x - 4), so (x = c + 4), then (x - 2 = c + 2) and (x - 6 = c - 2), which is the same substitution as before.So, I think my initial approach is correct, and the solutions are as I derived.Therefore, the solutions are:(x = 4 pm isqrt{12 - 2sqrt{30}}) and (x = 4 pm isqrt{12 + 2sqrt{30}}).Alternatively, simplifying (sqrt{12 pm 2sqrt{30}}) as (sqrt{6 pm sqrt{6}} pm sqrt{6 mp sqrt{6}}), but that might not be necessary.Alternatively, perhaps I can rationalize or simplify further, but I think this is as simplified as it gets.So, in conclusion, the solutions to the equation [(x - 2)^4 + (x - 6)^4 = -16] are:(x = 4 pm isqrt{12 - 2sqrt{30}}) and (x = 4 pm isqrt{12 + 2sqrt{30}}).But wait, earlier I thought the solutions were (4 pm 2isqrt{3}) and (4 pm isqrt{2}), but that was based on an incorrect factorization. So, the correct solutions are the ones I just derived.Alternatively, perhaps I can write (sqrt{12 pm 2sqrt{30}}) in terms of simpler radicals, but I don't think it's possible. So, I'll stick with the solutions as they are.Therefore, the solutions are:(x = 4 pm isqrt{12 - 2sqrt{30}}) and (x = 4 pm isqrt{12 + 2sqrt{30}}).I think that's the final answer.

💡Okay, so I have this problem where I need to figure out how many ways I can choose 5 numbers from the set {1, 2, ..., 18} such that no two of these numbers are consecutive. Hmm, that sounds a bit tricky, but let me try to break it down.First, I know that when choosing numbers without any restrictions, the number of ways to choose 5 numbers from 18 is just the combination C(18,5). But here, there's an added condition: no two numbers can be next to each other. So, I need to adjust my approach to account for that.Let me think about what it means for numbers to be non-consecutive. If I pick a number, say 3, then I can't pick 2 or 4. So, each number I pick kind of "blocks" the numbers immediately before and after it. But since I'm choosing multiple numbers, I have to make sure that none of them are adjacent to each other.Maybe I can model this problem by considering the positions of the chosen numbers and the gaps between them. If I have 5 numbers, there will be 4 gaps between them. Each of these gaps must be at least 1 number to ensure that the numbers aren't consecutive. So, if I think of the numbers as objects and the gaps as spaces between them, I can use some combinatorial method to count the number of valid arrangements.Wait, I remember something about stars and bars. Is that applicable here? Let me recall. The stars and bars theorem is used to find the number of ways to distribute identical objects into distinct bins. In this case, maybe I can think of the numbers as stars and the gaps as bars.But I need to adjust for the fact that the numbers are in a specific range from 1 to 18. So, perhaps I can transform the problem into a different space where the non-consecutiveness condition is automatically satisfied.Let me try this: suppose I choose 5 numbers a1, a2, a3, a4, a5 such that 1 ≤ a1 < a2 < a3 < a4 < a5 ≤ 18 and no two ai's are consecutive. To ensure they are not consecutive, I can define new variables bi where bi = ai - (i-1). This transformation shifts each number by its position, effectively creating gaps between them.So, if I define b1 = a1, b2 = a2 - 1, b3 = a3 - 2, b4 = a4 - 3, b5 = a5 - 4, then the new variables b1, b2, b3, b4, b5 will satisfy 1 ≤ b1 < b2 < b3 < b4 < b5 ≤ 18 - 4 = 14. Because each subsequent number is shifted by one more than the previous, ensuring that the gaps are at least 1.Therefore, the problem reduces to choosing 5 numbers from 1 to 14 without any restrictions, which is simply the combination C(14,5). Let me calculate that.C(14,5) is calculated as 14! / (5! * (14-5)!) = 14! / (5! * 9!) = (14 × 13 × 12 × 11 × 10) / (5 × 4 × 3 × 2 × 1) = (240240) / (120) = 2002.So, there are 2002 ways to choose 5 non-consecutive numbers from 1 to 18.Wait, let me double-check this reasoning. By transforming the original problem into selecting numbers from a smaller set, I ensure that the non-consecutiveness is maintained. The transformation effectively creates the necessary gaps, so the count should be accurate. Yeah, I think this makes sense.Another way to think about it is by considering the positions of the chosen numbers and the required gaps. If I have 5 numbers, they occupy 5 positions, and to ensure no two are consecutive, there must be at least one number between each pair. So, the minimum number of positions required is 5 + 4 = 9. Since we're choosing from 18 numbers, the remaining 18 - 9 = 9 numbers can be distributed as additional gaps. This is similar to placing 5 objects with at least one space between them and distributing the remaining spaces.Using the stars and bars method, the number of ways to distribute these 9 extra gaps into the 6 possible gaps (before the first number, between each pair, and after the last number) is C(9 + 6 -1, 6 -1) = C(14,5), which is the same result as before. So, this confirms that the answer is indeed 2002.I think I've covered the problem from different angles and both methods lead to the same conclusion. It seems solid. I don't see any flaws in the reasoning, so I'm confident that 2002 is the correct number of ways.

💡First, I need to determine the total number of ways Kim can select 2 shoes out of the 6 available. This can be calculated using the combination formula, which is:[text{Total combinations} = binom{6}{2} = frac{6!}{2!(6-2)!} = frac{6 times 5}{2 times 1} = 15]Next, I'll calculate the number of ways Kim can select 2 shoes of the same color. Since there are 3 pairs of shoes, each pair consisting of 2 shoes of the same color, there are 3 possible ways to select a matching pair:- Selecting both shoes from Pair A- Selecting both shoes from Pair B- Selecting both shoes from Pair CSo, there are 3 favorable outcomes.Finally, to find the probability of selecting 2 shoes of the same color, I'll divide the number of favorable outcomes by the total number of possible outcomes:[text{Probability} = frac{3}{15} = frac{1}{5}]Therefore, the probability that Kim will select 2 shoes of the same color is (boxed{(frac{1}{5})}).

💡First, I recognize that tossing three coins simultaneously results in (2^3 = 8) possible outcomes.Next, I identify the complementary event of getting at least one head, which is getting all tails. There is only 1 outcome where all three coins show tails.Using the complementary probability approach, the probability of getting at least one head is calculated as:[1 - frac{text{Number of all tails outcomes}}{text{Total number of outcomes}} = 1 - frac{1}{8} = frac{7}{8}]Therefore, the probability that at least one coin shows heads when tossing three coins is (boxed{frac{7}{8}}).

💡First, I need to establish the ratios between the number of fans for each team. The ratio of NY Yankees fans to NY Mets fans is 3:2, and the ratio of NY Mets fans to Boston Red Sox fans is 4:5.To combine these ratios, I'll make the number of NY Mets fans consistent in both ratios. I'll multiply the first ratio by 4 and the second ratio by 2:- Yankees : Mets = 3:2 becomes 12:8- Mets : Red Sox = 4:5 becomes 8:10Now, the combined ratio is Yankees : Mets : Red Sox = 12:8:10.Next, I'll calculate the total number of ratio units: 12 + 8 + 10 = 30.Given that there are 360 baseball fans in total, I'll determine the value of one ratio unit:30 units = 360 fans1 unit = 360 / 30 = 12 fansFinally, I'll find the number of NY Mets fans by multiplying the Mets' ratio units by the value of one unit:Mets fans = 8 units * 12 fans/unit = 96 fans

💡First, I need to factorize 12,000 into its prime components. I know that 12,000 can be broken down into smaller parts that are easier to work with. Starting with 12,000, I can express it as 120 multiplied by 100. Then, breaking down 120 into 12 and 10, and further breaking down 12 into 2 squared times 3, and 10 into 2 times 5. Finally, 100 is 10 cubed, which is (2 times 5) cubed. Putting it all together, 12,000 equals 2 raised to the 5th power times 3 times 5 raised to the 3rd power.Next, to find the factors of 12,000 that are perfect squares, I need to consider the exponents of each prime factor. A perfect square has even exponents in its prime factorization. For the prime number 2, which has an exponent of 5, the possible even exponents are 0, 2, and 4. For the prime number 3, since its exponent is 1, the only possible even exponent is 0. For the prime number 5, with an exponent of 3, the possible even exponents are 0 and 2.Now, I can count the combinations of these exponents to determine the number of perfect square factors. There are three choices for the exponent of 2 (0, 2, 4), one choice for the exponent of 3 (0), and two choices for the exponent of 5 (0, 2). Multiplying these together, I get 3 times 1 times 2, which equals 6.Therefore, there are 6 factors of 12,000 that are perfect squares.