We can use combination method,

(3c1) * (4c1)^4 > any of answer choices

None of above.

Answer [spoiler]{E}[/spoiler]?

The intent of the problem seems to be as follows:

parulmahajan89 wrote:A computer game has five difficulty levels. Each level offers a choice of four different scenarios, except for the first level, which offers a choice of only three different scenarios. For each game, a player must select one level and one scenario. How many different games could a player select?

a) 18
b) 19
c) 20
d) 21
e) 22

First level:
Since 3 scenarios are available, the number of options = 3.

Remaining 4 levels:
Since there 4 scenarios for each of these 4 levels, the number of options = 4*4 = 16.

Total games = 3+16 = 19.

The correct answer is B.

Another approach is to WRITE IT OUT.
Let the 5 levels be A, B, C, D and E.

For A, there are 3 scenarios available:
A1, A2, A3 --> 3 options.

For each of B, C, D, E, there are 4 scenarios available:
B1, B2, B3, B4 --> 4 options.
C1, C2, C3, C4 --> 4 options.
D1, D2, D3, D4 --> 4 options.
E1, E2, E3, E4 --> 4 options.

Total games = 3+4+4+4+4 = 19.

The correct answer is B. I am not sure about the source.

parulmahajan89 wrote:The correct answer is B. I am not sure about the source.

Is this a GMAT Question ? If this is not a GMAT question then what's the source ?