Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appears, is

We have to count all possible six-digit numbers that are formed with the digits $$1,3,5,7,9$$ under the two simultaneous conditions:

(i) no digit outside this set can occur, and

(ii) each of these five digits must appear at least once in the six positions.

Because we need six places but only five distinct digits, exactly one of the five digits will appear twice and every other digit will appear once.

First we decide which digit is to be repeated. There are $$5$$ choices for this.

After choosing the repeated digit, we decide in which two of the six positions this repeated digit will sit. The number of ways to choose these two positions is given by the combination formula $$ {}^{6}C_{2}=\frac{6!}{2!\,4!}. $$

Once those two places are fixed, the remaining four different digits must fill the remaining four places. Because all four digits are distinct, they can be arranged in $$ 4! $$ ways.

Multiplying the independent choices we get the total required count:

$$ \text{Total} \;=\; 5 \times {}^{6}C_{2} \times 4!. $$

Substituting the values of the factorials, we calculate step by step:

$$ {}^{6}C_{2}=\frac{6!}{2!\,4!}= \frac{720}{2\times24}=15, $$

and

$$ 4!=24. $$

So

$$ \text{Total}=5 \times 15 \times 24. $$

Now multiply sequentially:

$$ 15 \times 24 = 360, $$

and

$$ 5 \times 360 = 1800. $$

For comparison with the given options we rewrite the result using $$6! = 720$$:

$$ 1800 = \frac{5}{2}\times 6!. $$

Hence, the correct answer is Option D.