answer:1. Since this is a step-by-step counting problem, we first select 3 male athletes, which can be done in binom{6}{3} ways. Then we select 2 female athletes, which can be done in binom{4}{2} ways. So, in total, there are binom{6}{3} times binom{4}{2} = 20 times 6 = 120 methods of selection. boxed{120} 2. The selection method of "only male captain" can be carried out in binom{8}{4} ways since we have 8 other athletes to choose from. The same is valid for "only female captain". The method where "both male and female captains are selected" has binom{8}{3} ways as we are left with three spots to fill after selecting both captains. Therefore, in total, there are 2 times binom{8}{4} + binom{8}{3} = 2 times 70 + 56 = 196 methods. boxed{196} 3. The opposite event of "at least one female athlete" is "all male athletes." From the 10 athletes, selecting 5 can be done in binom{10}{5} ways. Among these, the ones with all male athletes are in binom{6}{5} ways. Thus, the methods for having "at least one female athlete" are binom{10}{5} - binom{6}{5} = 252 - 6 = 246. boxed{246} 4. When the female captain is included, the remaining selection can be any of the remaining 9 athletes, so there are binom{9}{4} ways. When the female captain is not selected, the male captain must be chosen, and the remainder can then be chosen from the 8 remaining athletes, leading to binom{8}{4} ways. However, from these binom{8}{4} ways, we have to subtract the selections without any female athletes, which is binom{5}{4} ways. Therefore, the total number of methods when both a captain and at least one female athlete must be present is binom{9}{4} + binom{8}{4} - binom{5}{4} = 126 + 70 - 5 = 191. boxed{191}

answer:Since all terms in the arithmetic sequence {a_n} are non-zero and satisfy a_3 - frac{a_7^2}{2} + a_{11} = 0, By the properties of arithmetic sequences, we have a_3 + a_{11} = 2a_7. So, 4a_7 - a_7^2 = 0. Solving this equation, we get a_7 = 4 (and a_7 = 0, which is discarded since all terms are non-zero). Hence, b_7 = a_7 = 4. Therefore, b_1 cdot b_{13} = a_7^2 = boxed{16}.

answer:**Analysis** This question examines the general term formula of a sequence, which is a basic problem. **Solution** The sequence sqrt{2}, sqrt{5}, 2sqrt{2}, sqrt{11}, ldots can be rewritten as sqrt{2}, sqrt{5}, sqrt{8}, sqrt{11}, ldots Its general term formula is a_n= sqrt{3n-1}, Therefore, the correct choice is boxed{text{B}}.

answer:1. Let's denote the probability of Ostap winning against the Champion of Russia, the Champion of Vasyukov, and the Champion of the World as (p_1), (p_2), and (p_3) respectively. 2. According to the problem, Ostap needs to win two games in a row to become the Absolute Champion. We are to determine the most favorable sequence of opponents. 3. We assume that Ostap's chances of beating the Champion of Russia are higher than beating the Champion of the World but lower than beating the Champion of Vasyukov. 4. Hence, we can conclude the probabilities are ordered as (p_3 < p_1 < p_2). 5. Let's calculate the probability of Ostap winning two consecutive matches in different orders of playing: # Step-by-step: - **Scenario 1**: Play against Champion of Russia, then Champion of Vasyukov, then Champion of the World. [ text{Probability} = p_1 times p_2 + left(1 - p_1right) times p_2 times p_3 ] [ = p_1 times p_2 + p_2 left(1 - p_1right) times p_3 ] [ = p_1 times p_2 + p_2 times p_3 - p_1 times p_2 times p_3 ] [ = p_2 left(p_1 + p_3 - p_1 times p_3right) ] - **Scenario 2**: Play against Champion of Vasyukov, then Champion of Russia, then Champion of the World. [ text{Probability} = p_2 times p_1 + left(1 - p_2right) times p_1 times p_3 ] [ = p_2 times p_1 + p_1 times p_3 - p_2 times p_1 times p_3 ] [ = p_1 left(p_2 + p_3 - p_2 times p_3right) ] 6. To compare these probabilities, we need to check when: [ p_2 left(p_1 + p_3 - p_1 times p_3right) > p_1 left(p_2 + p_3 - p_2 times p_3right) ] 7. Simplify the inequality, [ p_2 left(p_1 + p_3right) - p_1 times p_2 times p_3 > p_1 left(p_2 + p_3right) - p_2 times p_1 times p_3 ] [ p_2 times p_1 + p_2 times p_3 > p_1 times p_2 + p_1 times p_3 ] [ p_2 times p_3 > p_1 times p_3 ] [ p_2 > p_1 ] 8. Given that (p_2 > p_1) is satisfied (since (p_1) is the probability after Vasyukov), this inequality is true. # Conclusion: Since (p_2 > p_1), the more suitable strategy is that the second opponent should be the weakest within the sequence. Hence, Ostap should choose the Champion of Vasyukov as the second opponent. The order of first and third opponents does not matter once the second opponent is determined. [ boxed{text{Champion of Vasyukov as the second opponent, order of first and third does not matter}} ]