A seven-digit number comprises of only 2's and 3's. How many of these are…

Number should be a multiple of 3 and 4. So, the sum of the digits should be a multiple of 3. We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3.

(The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

All seven 3's - No possibility.

Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order.

No of possibilities = 5!/3!2! = 10

Six 2's and one 3 - The first 5 digits should all be 2's. So, there is only one number 2222232.

So, there are a total of 10 + 1 = 11 solutions.

The question is "A seven-digit number comprises of only 2's and 3's. How many of these are multiples of 12?"

Hence the answer is "11"

Choice A is the correct answer.