Math Problem Statement

1から7までの整数から異なる4個を選んで4桁の整数をつくるとき、全体で何通りできるか。そのうち奇数は何通りか。また、各位が小さい順に並んでいる数は何通りか。

Solution

The problem is asking about selecting 4 different integers from the set {1, 2, 3, 4, 5, 6, 7} to form four-digit numbers. It has the following parts:

  1. The total number of possible four-digit numbers that can be made by selecting 4 different integers from the given set.
  2. The number of those numbers that are odd.
  3. The number of those numbers where the digits are arranged in increasing order.

Let's break it down step by step:

1. Total number of possible four-digit numbers ("アイウ" box):

We are selecting 4 digits from the set {1, 2, 3, 4, 5, 6, 7}. The number of ways to select 4 digits out of 7 is:

(74)=7×6×5×44×3×2×1=35\binom{7}{4} = \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35

After selecting the digits, they can be arranged in 4!4! (4 factorial) different ways. Therefore, the total number of possible four-digit numbers is:

35×4!=35×24=84035 \times 4! = 35 \times 24 = 840

So, the total number of possible four-digit numbers is 840.

2. The number of odd numbers ("キクケ" box):

To form an odd number, the last digit must be odd. The odd numbers in the set are {1, 3, 5, 7}. So, the last digit must be one of these four.

After choosing the last digit, we are left with 6 digits, from which we need to select 3 digits. The number of ways to select 3 digits from the remaining 6 is:

(63)=6×5×43×2×1=20\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20

After selecting the 3 digits, they can be arranged in any order in the first three positions, which is 3!3! ways. Therefore, the total number of odd numbers is:

4×20×6=4804 \times 20 \times 6 = 480

So, the number of odd numbers is 480.

3. The number of numbers where the digits are in increasing order ("コサ" box):

For a number to have its digits in increasing order, the 4 digits must be selected, and their arrangement must automatically follow the increasing order. For example, if we select {2, 3, 5, 7}, the only possible arrangement is 2357.

The number of ways to select 4 digits from 7 and arrange them in increasing order is just the number of ways to select 4 digits from 7:

(74)=35\binom{7}{4} = 35

So, the number of numbers with digits in increasing order is 35.

Final Answers:

  • Total number of four-digit numbers (アイウ): 840
  • Number of odd numbers (キクケ): 480
  • Number of numbers with digits in increasing order (コサ): 35

Would you like a detailed breakdown of any specific step, or have additional questions? Here are some related questions for further practice:

  1. How many even numbers can be formed instead of odd numbers?
  2. What if the digits were chosen from {1, 2, 3, 4, 5, 6} instead of {1, 2, 3, 4, 5, 6, 7}?
  3. How would the solution change if repeated digits were allowed?
  4. What is the probability of randomly selecting an odd number from all possible four-digit numbers?
  5. How many numbers can be formed if no digits are allowed to repeat?

Tip: Always start by identifying restrictions (e.g., odd or even number, ascending order) to simplify the calculation.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Factorials
Odd and Even Numbers
Permutations

Formulas

Combination formula: C(n, r) = n! / (r!(n-r)!)
Permutation formula: P(n, r) = n! / (n-r)!
Factorial formula: n! = n × (n-1) × ... × 1

Theorems

Basic Counting Principle

Suitable Grade Level

Grades 10-12