Question

Помогите решить задачу!! Пожалуйста!математика 6 класс На районной олимпиаде по математике оказалось шесть победителей.Однако на областнуюолимпиаду можно отправить только двоих. а) сколько существует вариантов выбора двух кандидатов? УКАЗАНИЕ : ДАЙТЕ КАЖДОМУ ПОБЕДИТЕЛЮ НОМЕР - ОТ 1 ДО 6. б) Сколько существует вариантов , если один из шести ребят признан лучшим и он обязательно будет участвовать в областной олимпиаде

Answer 1

А): с первым участником: 1 и 2, 1 и 3, 1 и 4, 1 и 5, 1 и 6,
со вторым: 2 и 3, 2 и 4, 2 и 5, 2 и 6
с третьим: 3 и 4, 3 и 5, 3 и 6
с четвертым: 4 и 5, 4 и 6
с пятым: 5 и 6
с шестым уже все есть,итого = 15 вариантов

б)6-1= 5 следовательно и вариантов существует тоже 5

Answer 2

Problem Analysis

We are given that there are six winners in a district-level math Olympiad, but only two of them can be sent to the regional-level Olympiad. We need to determine the number of possible combinations for selecting two candidates. Additionally, we need to calculate the number of combinations if one of the six winners is already selected to participate in the regional-level Olympiad.

Solution

a) To find the number of ways to choose two candidates out of six, we can use the concept of combinations. The formula for combinations is given by:

nCr = n! / (r! * (n-r)!)

Where n is the total number of candidates and r is the number of candidates to be selected.

In this case, we have n = 6 (total number of candidates) and r = 2 (number of candidates to be selected). Plugging these values into the formula, we get:

6C2 = 6! / (2! * (6-2)!)

Simplifying the expression:

6C2 = 6! / (2! * 4!) = (6 * 5 * 4!) / (2! * 4!) = (6 * 5) / (2 * 1) = 15

Therefore, there are 15 possible combinations to choose two candidates out of six.

b) If one of the six winners is already selected to participate in the regional-level Olympiad, we need to choose one candidate from the remaining five winners to accompany the selected candidate. The number of ways to choose one candidate out of five can be calculated using the same formula as before:

5C1 = 5! / (1! * (5-1)!) = 5

Therefore, if one of the six winners is already selected, there are 5 possible combinations to choose the second candidate.

Conclusion

a) There are 15 possible combinations to choose two candidates out of six winners. b) If one of the six winners is already selected, there are 5 possible combinations to choose the second candidate.

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