Question

The numbers can be formed using the digits $1, 2, 3, 4, 3, 2, 1$ so that odd digits always occupy odd places in ways.

• A. 9
• B. 18
• C. 6
• D. 3

Hint:

Always match the counts of specific slots to the counts of item types first. Since the number of odd positions exactly matches the number of odd digits, the problem cleanly splits into two fully independent sub-problems: permuting $\{1,1,3,3\}$ and permuting $\{2,2,4\}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must count the distinct 7-digit permutations of {1,1,2,2,3,3,4} such that odd digits occupy odd-indexed positions and even digits occupy even-indexed positions.

Step 2: Key Formula or Approach:
The total arrangements equal the permutations of odd digits in the 4 odd slots multiplied by the permutations of even digits in the 3 even slots, accounting for repetitions.

Step 3: Detailed Explanation:
Odd positions: 4 slots for {1,1,3,3} → 4!/(2!2!) = 6 ways. Even positions: 3 slots for {2,2,4} → 3!/2! = 3 ways. Total permutations = 6 × 3 = 18 ways.

Step 4: Final Answer:
The total number of ways is 18, matching option (B).