answer:To solve this problem, let's break it down into a step-by-step solution: 1. **Label the Shoes**: We label the left shoes as L_1, dots, L_{10} and the right shoes as R_1, dots, R_{10}. This gives us a clear way to identify each shoe and its pair. 2. **Total Pairings**: There are 10! possible ways to pair the shoes since each left shoe can be paired with any right shoe. 3. **Defining Bad Pairings**: A pairing is considered "bad" if there exists a collection of k<5 pairs that includes both the left and right shoes of k adults. This means we need to find pairings that do not form such collections. 4. **Finding Bad Collections**: To find a bad collection, we can start with any right shoe, say R_1, and follow its pairing to a left shoe, say L_i. Since the pairing is bad, R_i must also be in the collection. We continue this process, moving from right shoe to its paired left shoe and then to the corresponding right shoe, until we return to R_1. This forms a cycle. 5. **Cycle Notation**: The process described forms a cycle in permutation terms. A bad pairing occurs if there is a cycle with a length of less than 5. Therefore, we need to count pairings where every cycle has a length of at least 5. 6. **Counting Valid Pairings**: - **Single Cycle of Length 10**: This case yields 9! valid pairings because we can arrange the remaining 9 shoes in any order after choosing the first pair. - **Two Cycles of Length 5**: This case yields frac{{10choose 5}}{2} cdot {4!}^2 pairings. We choose 5 pairs out of 10, divide by 2 to account for the indistinguishability of the cycles, and then arrange the shoes within each cycle, yielding {4!}^2 arrangements. 7. **Calculating the Probability**: The probability of not forming a bad pairing is the sum of the probabilities of the two valid cases out of the total 10! pairings. [ frac{9! + frac{{10choose 5}}{2} cdot {4!}^2}{10!} = frac{1}{10} + frac{1}{50} = frac{3}{25} ] 8. **Final Answer**: Therefore, the probability that no collection of k<5 pairs made by the child contains the shoes from exactly k of the adults is frac{3}{25}. Since we need to find m+n where the probability is frac{m}{n} and m and n are relatively prime, we have m=3 and n=25. Thus, m+n=3+25=boxed{28}.

answer:Given the problem statement, we are required to determine all functions ( f: mathbb{R} rightarrow mathbb{R} ) such that for any ( x, y in mathbb{R} ), the following conditions hold: [ fleft(x^2 + 2xy + y^2right) = (x+y)[f(x) + f(y)] tag{1} ] and [ |f(x) - kx| leq |x^2 - x|. tag{2} ] 1. **Substitute ( x = 0 ) and ( y = 0 ) in equation (1):** [ f[0^2 + 2 cdot 0 cdot 0 + 0^2] = (0 + 0)[f(0) + f(0)] ] [ f(0) = 0. tag{3} ] 2. **Substitute ( y = 0 ) in equation (1) and use equation (3):** [ fleft(x^2 + 2 cdot x cdot 0 + 0^2right) = (x + 0)[f(x) + f(0)] ] [ f(x^2) = x f(x). tag{4} ] 3. **Substitute ( x = -x ) in equation (4):** [ f[(-x)^2] = -x f(-x) ] [ f(x^2) = -x f(-x). ] Combining with ( f(x^2) = x f(x) ), we get: [ x f(x) = -x f(-x). ] Therefore, [ f(-x) = -f(x) quad (text{for} x neq 0). tag{5} ] 4. Combining equations (3) and (5), it follows that: [ f(-x) = -f(x). ] Thus, ( f(x) ) is an odd function. 5. **Using equation (4), we have:** [ frac{f(x^2)}{x^2} = frac{f(x)}{x}. ] From this, for any ( x > 0 ), [ frac{f(x)}{x} = frac{f(x^{1/2})}{x^{1/4}} = frac{f(x^{1/4})}{x^{1/8}} = cdots = frac{f(x^{1/2^n})}{x^{1/2^n}} quad text{for} n in mathbb{N}^+. ] 6. **Utilize equation (2) with ( x = x^{1/2^n} ):** [ left| frac{f(x)}{x} - k right| leq left| x^{1/2^n} - 1 right|. ] As ( n rightarrow +infty ), ( x^{1/2^n} rightarrow 1 ): [ left| frac{f(x)}{x} - k right| rightarrow 0. ] Hence, [ frac{f(x)}{x} = k. ] From which it follows that [ f(x) = kx quad text{for} x > 0. ] 7. **Considering the oddness of ( f(x) ) from (5), we conclude:** [ f(x) = kx quad text{for all} x in mathbb{R}. ] Finally, we verify that ( f(x) = kx ) satisfies all conditions given in the problem statement. Hence, the function solution is: [ boxed{f(x) = kx text{ for all } x in mathbb{R}.} ]

answer:From the given condition S_3=11a_6, we have therefore 3a_1+3d=11(a_1+5d), therefore 2a_1+13d=0, which implies a_1+a_{14}=0, therefore a_7+a_8=0. Since the common difference of the arithmetic sequence is less than 0, S_n reaches its maximum value when n=7. Therefore, the answer is: boxed{B}. From the condition S_3=11a_6, using the general term formula, we can derive 3a_1+3d=11(a_1+5d), which gives 2a_1+13d=0, and a_7+a_8=0. Given that the common difference of the arithmetic sequence is less than 0, we can conclude the result. This problem tests the general term formula and sum formula of an arithmetic sequence and its monotonicity, examining reasoning and computational skills, and is considered a medium-level question.

answer:To solve the problem (-1)^1+(-1)^2+cdots+(-1)^{2006}, we observe the behavior of (-1)^k for different values of k: - When k is even, (-1)^k = 1. - When k is odd, (-1)^k = -1. Given this, we can pair each term in the sequence where an odd exponent is immediately followed by an even exponent. This gives us pairs of (-1 + 1) throughout the sequence: [ (-1)^1 + (-1)^2 + cdots + (-1)^{2005} + (-1)^{2006} = (-1 + 1) + (-1 + 1) + cdots + (-1 + 1). ] Each pair (-1 + 1) simplifies to 0. Since there are 1003 pairs (because there are 2006 terms, and each pair consists of 2 terms), the sum of all these pairs is: [ 0 + 0 + cdots + 0 = 0. ] Therefore, the sum (-1)^1+(-1)^2+cdots+(-1)^{2006} equals boxed{0}.