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 are given a collection of 7 digits: $\{1, 1, 2, 2, 3, 3, 4\}$. We need to find the number of unique 7-digit permutations such that the odd digits always fill the odd-indexed position slots.

Step 2: Key Formula or Approach:
The total number of positions is 7, indexed from 1 to 7: $$\text{Odd places: } \{1, 3, 5, 7\} \quad (\text{4 positions})$$ $$\text{Even places: } \{2, 4, 6\} \quad (\text{3 positions})$$ Let's categorize our given digits: $$\text{Odd digits: } \{1, 1, 3, 3\} \quad (\text{4 numbers})$$ $$\text{Even digits: } \{2, 2, 4\} \quad (\text{3 numbers})$$ Since there are exactly 4 odd positions for the 4 odd digits, the odd digits must completely fill the odd positions, leaving the 3 even digits to completely fill the 3 even positions. We compute the permutations of each group with repetitions using $\frac{n!}{p! \cdot q!}$.

Step 3: Detailed Explanation:
Let's find the number of distinct ways to arrange the odd digits $\{1, 1, 3, 3\}$ into the 4 odd position slots ($1^{\text{st}}, 3^{\text{rd}}, 5^{\text{th}}, 7^{\text{th}}$): $$\text{Ways (Odd)} = \frac{4!}{2! \cdot 2!} = \frac{24}{2 \cdot 2} = 6 \text{ ways}$$ Next, let's find the number of distinct ways to arrange the even digits $\{2, 2, 4\}$ into the 3 even position slots ($2^{\text{nd}}, 4^{\text{th}}, 6^{\text{th}}$): $$\text{Ways (Even)} = \frac{3!}{2!} = \frac{6}{2} = 3 \text{ ways}$$ By the fundamental counting principle, the total number of valid permutations is the product of the independent arrangements of the odd and even subsets: $$\text{Total valid ways} = \text{Ways (Odd)} \times \text{Ways (Even)} = 6 \times 3 = 18 \text{ ways}$$

Step 4: Final Answer:
The total number of ways to form such numbers is 18, which corresponds to option (B).