Question

РАБОТА В ГРУППЕ 2 Реши задачу.В клетке находились желтый, зеленый, голубой и синий попу-гаи. Айне разрешили выбрать двух попугаев. Сколько у Айнывариантов выбора?PAEOTA BIAPE3Выполни задание.Сколько существует чисел, больших, чем 3 528, каждое из ко-торых можно получить перестановкой цифр данного числа?САМОСТОЯТЕЛЬНАЯ РАБОТА4 Реши задачу, .В зоомагазине Арману понравились 5 видоврыбок. Цена рыбок одинакова, но денег хва-тит только на 3 рыбки. Сколько вариантов вы-бора трех рыбок из пяти видов есть у Армана?ПОЖАЛУЙСТА ОБЪЯСНИТЕ КАК УЧИТЕЛЬ ЧТО НАДО ДЕЛАТЬ СНАЧАЛО​

Answer 1

Ответ:4,2,3

Пошаговое объяснение:

Изначально в 4 говорят о рыбках а продолжения нет,поэтому в начале 4.Потом второй ведь продолжается про зоомагазин.В итоге остаётся 3.

Answer 2

Ответ:

Пошаговое объяснение:

Задание 2.

ЖЗ, ЖГ, ЖС, ЗГ, ЗС, ГС

Ответ: 6 вариантов выбора

Задание 3.

3582, 5328, 5382, 5238, 5283, 5832, 5823, 8352, 8325, 8235, 8253, 8523, 8532

Ответ: 13 вариантов чисел

Задание 4

Обозначим рыбок по номерам: 12345

123, 124, 125, 134, 135, 145, 234, 235, 245

Ответ: 9 вариантов

Answer 3

Problem 1: Choosing Two Parrots

In the first problem, Ayna is allowed to choose two parrots from a group of yellow, green, blue, and blue parrots. We need to determine how many options Ayna has for choosing the parrots.

To solve this problem, we can use the concept of combinations. The number of ways to choose two parrots from a group of four can be calculated using the formula for combinations:

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

Where n is the total number of parrots and r is the number of parrots to be chosen.

In this case, n = 4 (since there are four parrots) and r = 2 (since Ayna needs to choose two parrots).

Plugging in the values, we get:

4C2 = 4! / (2!(4-2)!) = 4! / (2!2!) = (4 * 3 * 2 * 1) / ((2 * 1) * (2 * 1)) = 6

Therefore, Ayna has 6 options for choosing two parrots from the given group.

Problem 2: Permutations of Numbers Greater than 3528

In the second problem, we need to find the number of numbers greater than 3528 that can be obtained by permuting the digits of the given number.

To solve this problem, we can consider the different positions of the digits in the number and calculate the number of permutations.

The given number is 3528. To find the number of permutations, we need to consider the digits 3, 5, 2, and 8.

Since we are looking for numbers greater than 3528, the first digit can only be 5, 8, or 9. The remaining digits can be any permutation of the digits 3, 2, and 8.

Therefore, the number of permutations is:

3 (choices for the first digit) * 3! (permutations of the remaining digits) = 3 * 3! = 3 * 6 = 18

So, there are 18 numbers greater than 3528 that can be obtained by permuting the digits of the given number.

Problem 3: Choosing Three Fish from Five

In the third problem, Arman likes 5 types of fish in a pet store. He has enough money to buy only 3 fish. We need to determine how many options Arman has for choosing three fish from the five types available.

To solve this problem, we can again use the concept of combinations. The number of ways to choose three fish from a group of five can be calculated using the combination formula:

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

In this case, n = 5 (since there are five types of fish) and r = 3 (since Arman needs to choose three fish).

Plugging in the values, we get:

5C3 = 5! / (3!(5-3)!) = 5! / (3!2!) = (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (2 * 1)) = 10

Therefore, Arman has 10 options for choosing three fish from the five types available.

Teacher's Instructions

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