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…
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
2592
240
3032
2352
Solution
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
$2592-240=2352$