The problem in the image is written in Japanese and involves choosing 4 digits from the set {1, 2, 3, 4, 5, 6, 7} to form four-digit numbers. It requires calculating the total number of such numbers, the number of odd numbers greater than 3500, and the number of numbers in increasing order from left to right.

Step-by-step analysis:

1. Total number of 4-digit numbers:

   • We are selecting 4 digits from the numbers 1 to 7, and since order matters, it's a permutation problem.
   • The total number of permutations of 4 digits chosen from 7 is calculated as: $P(7, 4) = \frac{7!}{(7-4)!} = 7 \times 6 \times 5 \times 4 = 840$ So, the total number of 4-digit numbers is 840.

2. Odd numbers greater than 3500:

   • Odd numbers must end with one of the odd digits: {1, 3, 5, 7}.
   • To ensure the number is greater than 3500, the first digit must be at least 4.
   • We analyze possible choices for the first and last digits and calculate the remaining combinations.

3. Numbers in increasing order:

   • Numbers like 2357, where each digit is in increasing order, are combinations (not permutations) because order is fixed as increasing.
   • To form a number with increasing digits, we just need to choose 4 digits from the set {1, 2, 3, 4, 5, 6, 7}.
   • The number of ways to choose 4 digits from this set is: $C(7, 4) = \frac{7!}{4!(7-4)!} = 35$

Let me know if you'd like further details or clarification!

Here are some related questions to deepen your understanding:

1. What if the numbers formed must be divisible by 5 instead of being odd?
2. How would the solution change if the range of numbers were {1, 2, 3, 4, 5, 6, 7, 8}?
3. How many numbers would be possible if repetition of digits were allowed?
4. What would be the number of even numbers greater than 4000?
5. How would you calculate the sum of all the valid 4-digit numbers?

Tip: When dealing with permutations and combinations, always carefully distinguish whether order matters to avoid confusion.