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

As we have to make a six-digit number by using all the digits $1, 3, 5, 7, 9,$ hence exactly one of the digits is repeating.

For the digit to repeat we have ${}^5{C}_1=5$ choices.
And, six digits in which one is repeating can be arranged in $\frac{6!}{2!}$ ways.
Hence, total such numbers $=5\times \frac{6!}{2!}=\frac{5}{2}\left(6!\right).$