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

Step-by-step 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 4 6 6 6=2592 Now, total numbers when no repetition is aligned & mode = 4 4 3 2 1 _ When last digit is 3 + 2(3 4 3 2 1) _ When last digit is 5 or 7 & =96+144