Question

Задача на размещение и сочетание: Сколько трёхзначных чисел, кратных пяти, можно записать с помощью цифр 4,5,6,7,8,9 при условии, что в каждом числе цифры не повторяются. Пожалуйста помогите, ни как не пойму как её решить!!!!!!!!!!!!

Answer 1

Чтобы число было кратно 5, на конце данного числа должна быть цифра 5.
Следовательно, в числе остается 2 неизвестных цифры (т.е. **5)
Дано 5 цифр, следовательно, нам необходима формула размещения из 5 по 2:
5! / (5-2)! = 4*5 = 20
Ответ: 20 чисел.

Answer 2

Task Analysis

To solve this problem, we need to determine how many three-digit numbers can be formed using the digits 4, 5, 6, 7, 8, and 9, with the condition that the digits cannot be repeated and the number must be divisible by 5.

Solution

To find the solution, we can break down the problem into smaller steps:

Step 1: Determine the possible choices for each digit in the three-digit number. Step 2: Count the number of possible combinations. Step 3: Determine which combinations are divisible by 5.

Let's go through each step in detail.

Step 1: Determine the possible choices for each digit

Since the digits cannot be repeated, the first digit can be any of the six given digits (4, 5, 6, 7, 8, 9). The second digit can be any of the remaining five digits, and the third digit can be any of the remaining four digits.

Step 2: Count the number of possible combinations

To count the number of possible combinations, we multiply the number of choices for each digit. So, the total number of combinations is:

Number of combinations = Number of choices for the first digit * Number of choices for the second digit * Number of choices for the third digit

Number of combinations = 6 * 5 * 4

Step 3: Determine which combinations are divisible by 5

To determine which combinations are divisible by 5, we need to check if the last digit of each combination is either 5 or 0. Since the last digit can only be one of the four remaining digits (5, 6, 7, 8), we need to remove the combinations that end with 6, 7, or 8.

Final Answer

To summarize, the number of three-digit numbers that can be formed using the digits 4, 5, 6, 7, 8, and 9, with the condition that the digits cannot be repeated and the number must be divisible by 5, is:

Number of combinations = 6 * 5 * 4 = 120

However, we need to remove the combinations that end with 6, 7, or 8. Since there are 3 remaining digits (5, 7, 9) that can be used as the last digit, we multiply the number of combinations by 3.

Number of combinations = 120 * 3 = 360

Therefore, there are 360 three-digit numbers that meet the given conditions.

Example Combinations

Here are a few examples of the possible combinations:

- 495 - 549 - 594 - 945 - 954

Please note that these are just a few examples, and there are many more possible combinations.

I hope this helps! Let me know if you have any further questions.