Explanatory Answer

Method of solving this CAT Quantitative Ability Question from Number System question

Correct Answer : 50

To form a four-digit number using only the digits 1, 2, and 3, with both 2 and 3 appearing at least once, we can follow the given approach:

Step 1: Select the two positions for the digits 2 and 3.

We have 4 positions available for placing the digit 2.
We have 3 positions remaining for placing the digit 3.

Step 2: Select the remaining two digits from the set {1, 1, 2, 3}.

There are six possible combinations: (1, 1), (2, 2), (3, 3), (1, 2), (1, 3), or (2, 3).

Step 3: Arrange the four selected digits in different ways.

For the combinations (1, 1), (2, 2), and (3, 3): We have 4 positions to place the two identical digits, so the number of arrangements is given by 4!/(2!) = 12 for each combination.
For the combinations (1, 2), (1, 3), and (2, 3): We have 4 positions to place the two distinct digits, so the number of arrangements is given by 4! = 24 for each combination.

Step 4: Calculate the total number of possibilities.

Summing up the arrangements for each combination, we have:
12 + 12 + 12 + 24 + 24 + 24 = 108 arrangements.

However, upon reviewing the previous solution, there seems to be an error in the calculation. The correct total number of possibilities is 50, not 108.

To clarify further:

The arrangements (2, 3, 1, 1), (2, 3, 2, 2), (2, 3, 3, 3), (2, 3, 1, 2), (2, 3, 1, 3), and (2, 3, 2, 3) can be rearranged in 12, 4, 4, 12, 12, and 6 ways, respectively.
Summing up these arrangements, we have 12 + 4 + 4 + 12 + 12 + 6 = 50 possible arrangements.

Therefore, the total number of four-digit numbers that can be formed using only the digits 1, 2, and 3, with both 2 and 3 appearing at least once, is 50.