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question:a group of twelve people where each person shakes hands with exactly three others, find the number of distinct handshaking arrangements, modulo 1000. The arrangement is considered different if at least one pair of people who shake hands in one arrangement do not do so in another.
answer:Given that each person shakes hands with three others, we consider the graph theory representation with 'rings'. The possible configurations are: three rings of four, two rings of four and one ring of four, and one ring of eight. **Case 1: Three rings of four** - Choose groups of four: frac{binom{12}{4}binom{8}{4}binom{4}{4}}{3!} to account for the selection of groups without order. - Within a ring of four, there are frac{(4-1)!}{2} = 3 ways to arrange the people, considering rotations and reflections. - Total for this case: left(frac{binom{12}{4}binom{8}{4}binom{4}{4}}{3!}right) times 3^3 = 34650. **Case 2: One ring of four and one ring of eight** - Choose a group of four: binom{12}{4} and the remaining eight automatically form the other group. - Arrangements in the ring of four: frac{3!}{2} = 3. - Arrangements in the ring of eight: frac{7!}{2} = 2520. - Total for this case: binom{12}{4} times 3 times 2520 = 831600. Summing the cases, we get 34650 + 831600 = 866250. Thus, the remainder when N is divided by 1000 is boxed{250}.
question:f the equation x^2+2(a1)x+2a+6=0 has one positive and one negative real root, then find the range of the real number a.
answer:Let f(x) = x^2+2(a-1)x+2a+6. Since the equation x^2+2(a-1)x+2a+6=0 has one positive and one negative real root, it follows that f(0) = 2a+6 < 0. Solving this inequality gives a < -3. Therefore, the range of a is boxed{a < -3}.
question:Given the universal set U={1,2,3,4,5,6,7,8}, and the complement of set B with respect to U, denoted as C_{U}B={1,3}, then the set B is ( ) A: {2,4,5,6,7,8} B: {4,5,6,7,8} C: {2,5,6,7,8} D: {5,6,7,8}
answer:Since the universal set U={1,2,3,4,5,6,7,8}, and since C_{U}B={1,3}, therefore, B={2,4,5,6,7,8}, hence, the correct option is boxed{A}.
question:A rectangular yard measures 30 meters by 10 meters and contains two flower beds in the shape of congruent equilateral triangles. The remainder of the yard is trapezoidal. The lengths of the parallel sides of the trapezoid are 40 meters and 20 meters. What fraction of the yard is occupied by the flower beds? A) frac{5sqrt{3}}{30} B) frac{5}{30} C) frac{5sqrt{2}}{30} D) frac{5}{20}
answer:1. **Calculate the dimensions of the equilateral triangles:** Given that the parallel sides of the trapezoid are 40 meters and 20 meters, the difference between these sides is (40 - 20 = 20) meters. Assume each triangle takes up (10) meters from this difference. [ text{Side length of each triangle} = 10 text{ meters} ] 2. **Calculate the area of one equilateral triangle:** The area (A) of an equilateral triangle with side length (s) is given by: [ A = frac{sqrt{3}}{4} s^2 ] Substituting (s = 10) meters: [ A = frac{sqrt{3}}{4} times 10^2 = frac{sqrt{3}}{4} times 100 = 25sqrt{3} text{ square meters} ] 3. **Calculate the total area of the flower beds:** Since there are two such triangles: [ text{Total area of flower beds} = 2 times 25sqrt{3} = 50sqrt{3} text{ square meters} ] 4. **Calculate the area of the entire yard:** The yard has dimensions 30 meters by 10 meters: [ text{Area of the yard} = 30 times 10 = 300 text{ square meters} ] 5. **Determine the fraction of the yard occupied by the flower beds:** [ text{Fraction occupied by flower beds} = frac{text{Area of flower beds}}{text{Area of the yard}} = frac{50sqrt{3}}{300} = frac{5sqrt{3}}{30} ] Thus, the fraction of the yard occupied by the flower beds is frac{5sqrt{3}{30}}. The final answer is boxed{A}