Answer :
Final answer:
To find the number of natural numbers less than 7,000 with given digits, we calculate the combinations for each digit place and add the counts for one-, two-, three-, and four-digit numbers. After accounting for the inclusion of 0 which is not a natural number, we determine there are 526 possible numbers.
Explanation:
The question asks us to find the number of natural numbers less than 7,000 that can be formed using the digits 0, 1, 3, 7, 9, where repetition of digits is allowed. To form a four-digit number less than 7,000, the first digit can only be 0, 1, or 3. The other three digits can be any of the five given digits.
For the first digit, there are 3 choices (excluding 7 and 9 as they would create numbers exceeding 7,000). For each of the second, third, and fourth digits, there are 5 choices (as repetition is allowed). Therefore, the total number of possible combinations is the product of these choices.
Calculating this, we get: 3 (choices for the first digit) imes 5 (choices for the second digit) imes 5 (choices for the third digit) imes 5 (choices for the fourth digit) = 3 imes 5^3 = 375 possible four-digit numbers. Additionally, we can form one-, two-, and three-digit numbers as well, which would be 5^1 (for one-digit), 5^2 (for two-digit), and 5^3 (for three-digit) numbers.
Summing all possibilities gives us: 5 + 25 + 125 + 375 = 530 natural numbers.
However, as we noted that 0 is one of the choices for the first digit, this includes numbers such as 0, 00, 000, and 0000, which are not natural numbers. We have to subtract these cases from our total. Since there's only one zero in each category (one, two, three, and four-digit numbers), we subtract 4 from 530 to get the final count: 530 - 4 = 526.
