Question

Total number of 6-digit numbers in which all the odd digits and only odd digits appear, is

• A. 1,3,5,7,9.

Answer 1

1800

Answer 2

The problem requires forming 6-digit numbers using only odd digits {1, 3, 5, 7, 9}, with the condition that each of these five odd digits must appear at least once. Since the number has 6 digits and we must use 5 distinct digits, one of the odd digits must be repeated.

1. Choose the repeated digit: There are 5 options for the digit that will appear twice (1, 3, 5, 7, or 9).
2. Arrange the digits: Once a digit is chosen for repetition (e.g., if '1' is repeated, the digits are {1, 1, 3, 5, 7, 9}), we need to arrange these 6 digits to form a 6-digit number. The number of distinct permutations of these 6 digits is given by $\frac{6!}{2!}$. $\frac{6!}{2!} = \frac{720}{2} = 360$.
3. Total count: Since there are 5 choices for the repeated digit, the total number of such 6-digit numbers is $5 \times 360 = 1800$.