The number of different seven digit numbers that can be written using only three digits 1, 2 & 3 under the condition that the digit 2 occurs exactly twice in each number is-

Options

A. a. 2412
B. b. 120
C. c. 672
D. d. none of these

Answer

To solve the problem of finding the number of different seven-digit numbers that can be formed using the digits 1, 2, and 3, with the condition that the digit 2 occurs exactly twice, we can follow these steps:

### Step 1: Choose Positions for the Digit '2'
Since the digit '2' must occur exactly twice in the seven-digit number, we first need to choose 2 positions out of the 7 available positions for the digit '2'. The number of ways to choose 2 positions from 7 is given by the combination formula:

\[ \text{Number of ways to choose 2 positions} = \binom{7}{2} \]

Calculating this:

\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 2: Fill the Remaining Positions
After placing the digit '2' in 2 positions, we have 5 remaining positions to fill. The remaining digits that can be used are '1' and '3'. Each of the 5 positions can be filled with either '1' or '3'. Therefore, the number of ways to fill these 5 positions is:

\[ \text{Number of ways to fill 5 positions} = 2^5 \]

Calculating this:

\[ 2^5 = 32 \]

### Step 3: Calculate the Total Number of Combinations
Now, to find the total number of different seven-digit numbers that can be formed, we multiply the number of ways to choose the positions for '2' by the number of ways to fill the remaining positions:

\[ \text{Total combinations} = \binom{7}{2} \times 2^5 = 21 \times 32 \]

Calculating this:

\[ 21 \times 32 = 672 \]

### Conclusion
Thus, the total number of different seven-digit numbers that can be formed under the given conditions is **672**.

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