Total number of 6 - digit numbers in which only and all the five...

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 $\dfrac{6 !}{2 !}$ ways.
Hence, total such numbers $= 5 \times \dfrac{6 !}{2 !} = \dfrac{5}{2}\left(6 !\right) .$