How many numbers can be formed with the digits 1, 6, 7, 8, 6, 1 so that the odd digits always occupy the odd places.

Options

A. 15
B. 12
C. 18
D. 20

Answer

To solve the problem of how many numbers can be formed with the digits 1, 6, 7, 8, 6, 1 such that the odd digits always occupy the odd places, we can follow these steps:

### Step 1: Identify the digits and their types
The digits we have are: 1, 6, 7, 8, 6, 1.
- Odd digits: 1, 1, 7 (3 odd digits)
- Even digits: 6, 6, 8 (3 even digits)

### Step 2: Determine the positions for odd and even digits
In a 6-digit number, the positions are:
- Odd positions: 1st, 3rd, 5th
- Even positions: 2nd, 4th, 6th

### Step 3: Fill the odd positions with odd digits
We need to fill the 3 odd positions (1st, 3rd, 5th) with the 3 odd digits (1, 1, 7). Since the digit '1' is repeated, we will use the formula for permutations of a multiset:

\[ \text{Number of ways to arrange odd digits} = \frac{3!}{2!} = \frac{6}{2} = 3 \]

### Step 4: Fill the even positions with even digits
Next, we fill the 3 even positions (2nd, 4th, 6th) with the 3 even digits (6, 6, 8). Since the digit '6' is repeated, we will again use the formula for permutations of a multiset:

\[ \text{Number of ways to arrange even digits} = \frac{3!}{2!} = \frac{6}{2} = 3 \]

### Step 5: Calculate the total arrangements
The total arrangements can be calculated by multiplying the number of ways to arrange the odd digits and the even digits:

\[ \text{Total arrangements} = (\text{Ways to arrange odd digits}) \times (\text{Ways to arrange even digits}) = 3 \times 3 = 9 \]

### Step 6: Consider the arrangement of positions
Since the odd and even positions can be filled in two different ways (from left to right and right to left), we need to multiply the total arrangements by 2:

\[ \text{Final total arrangements} = 9 \times 2 = 18 \]

Thus, the total number of numbers that can be formed is **18**.

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