The number of 5 digit odd numbers greater than 40,000 that can be formed by using $3,4,5,6,7,0$ so that at least one of its digit must be repeated is

Since 5-digit numbers must be odd, their last digit is 3,5 , or 7 . We can choose the last digit any of 3 ways, as 3,5 or 7 We can then choose the first digit any of 4 ways, as 4,5 , 6 or 7 We choose the second digit as any of the 6 digits We choose the third digit as any of the 6 digits We choose the fourth digit as any of the 6 digits Total such numbers are $3 \times 4 \times 6 \times 6 \times 6=2592$ Now, total numbers when no repetition is $\begin{aligned} & \text { mode }=\underbrace{4 \times 4 \times 3 \times 2 \times 1}_{\text {When last digit is } 3}+\underbrace{2(3 \times 4 \times 3 \times 2 \times 1)}_{\text {When last digit is } 5 \text { or } 7} \\ & =96+144=240 \end{aligned}$
So, required number of 5 digit numbers are $2592-240=2352$