Solution:

Counting the number of possibilities can be done in a few steps.

First, we note that there are five choices for the first digit (1, 3, 5, 7, or 9). Then, there are four choices for the second digit (since we have used one of the digits already). Similarly, there are three choices for the third digit, two choices for the fourth digit, and one choice for the fifth digit.

Multiplying these numbers together, we find that there are 5! = 120 possible five-digit numbers that can be formed using the digits 1, 3, 5, 7, and 9.

However, this includes some numbers that have repeated digits. To count the number of possibilities that do not have repeated digits, we can use the principle of inclusion-exclusion.

There are ${5 \choose 1}$ = 5 ways to choose one of the digits to repeat twice. For each of these choices, there are ${4 \choose 1}$ = 4 ways to choose which of the remaining digits to use as the sixth digit.

Similarly, there are ${5 \choose 2}$ = 10 ways to choose two of the digits to repeat twice. For each of these choices, there are ${3 \choose 1}$ = 3 ways to choose which of the remaining digits to use as the sixth digit.

Continuing in this way, we find that there are

${5 \choose 1} \cdot {4 \choose 1} + {5 \choose 2} \cdot {3 \choose 1} + {5 \choose 3} \cdot {2 \choose 1} + {5 \choose 4} \cdot {1 \choose 1} + {5 \choose 5} \cdot {0 \choose 1} = 5 \cdot 4 + 10 \cdot 3 + 10 \cdot 2 + 5 \cdot 1 + 1 \cdot 0 = 50 + 30 + 20 + 5 + 0 = 105$

six-digit numbers that have no repeated digits and use only the digits 1, 3, 5, 7, and 9.

Therefore, the total number of six-digit numbers that can be formed using the digits 1, 3, 5, 7, and 9 such that each digit is used at least once is

120 - 105 = 15.

Since each of the five digits must be used at least once, we can choose which digit to repeat twice in ${5 \choose 1} = 5$ ways. Therefore, the total number of possibilities is

15 $\cdot$ 5 = 75.

However, we have counted each six-digit number with repeated digits twice (once for each way of choosing which digit to repeat). Therefore, the actual number of possibilities is

75/2 = 37.5,

which we round up to 38.

Therefore, there are 38 six-digit numbers that can be formed using the digits 1, 3, 5, 7, and 9 such that each digit is used at least once.

You are asking to form 6 digit number using 5 numbers and that too without repetition, please check the question.