Please specify the official answer.

Now there are 5 questions and we have to select 4. Now there are two ways of doing this.

Case 1. Select first two questions and last two from remaining 3.
Case 2. Select one question from first 2 and last 3.

Case 1: Number of ways of selecting = \(C_2^3\) = 3 Namely {1 2 3 4}{1 2 3 5}{1 2 4 5}
Case 2: Number of ways of selecting = \(C_1^2\) = 2 Namely {1 3 4 5}{2 3 4 5}

Now in Case 1 first two questions have 3 choices and last two questions have 4 choices. Hence total choices possible for case 1 = 3 * 3 * 4* 4 = 144.
Case 2 fist question has 3 choices and remaining has 4 choices Hence total choices possible for case 1 = 3 * 4 * 4* 4 = 192

Combining both we have = 144*3 + 192*2 = 816.

Hence option B.

PLEASE SHARE THE OFFICIAL ANSWER AS WELL AS THE SOURCE OF THE QUE.

PS: I don't think this is a GRE QUE