How many numbers can be formed using all the digits 1, 2, 3, 4, 3, 2, 1 such that even digits alwaysoccupy even plac

1. Break Down the Problem

We need to form numbers using the digits 1, 2, 3, 4, 3, 2, 1, with the condition that even digits (2, 2, 4) must only occupy even positions in the number.

2. Relevant Concepts

1. Total Digits: There are 7 total digits: 1, 2, 3, 4, 3, 2, 1.

2. Even and Odd Positions:

   • Even positions in a 7-digit number: 2nd, 4th, 6th (3 even positions).
   • Odd positions: 1st, 3rd, 5th, 7th (4 odd positions).

3. Arrangement of Digits:

   • Total arrangements of digits can be calculated using the formula for permutations of multiset:

   $\frac{n!}{n_1! \cdot n_2! \cdot \ldots \cdot n_k!}$

   where $n$ is the total number of items to arrange, and $n_1, n_2, \ldots, n_k$ are the counts of each distinct item.

3. Analysis and Detail

Step 1: Arrangement of Even Digits

• Even digits available: 2, 2, 4 (total of 3 digits).
• These need to be arranged in 3 even positions.

The number of ways to arrange the even digits is:

$\frac{3!}{2!} = 3$

Step 2: Arrangement of Odd Digits

• Odd digits available: 1, 1, 3, 3 (total of 4 digits).
• These can be arranged in the 4 odd positions.

The number of ways to arrange the odd digits is:

$\frac{4!}{2! \cdot 2!} = 6$

4. Verify and Summarize

Now, we find the total number of arrangements by multiplying the arrangements of even digits and odd digits:

$\text{Total arrangements} = (\text{Ways to arrange even digits}) \times (\text{Ways to arrange odd digits}) = 3 \times 6 = 18$

Final Answer

The total number of numbers that can be formed such that even digits always occupy even positions is 18.