Problem Solving

40. A computer game has five difficulty levels. In each level you can choose among four different scenarios except for the first level, where you can choose among three scenarios only. How many different games are possible?

a) 18
b) 19
c) 20
d) 21
e) None of the above

Can we use combination formulae here? and if yes we will get the right answer? I get that there are five game levels and 7 scenarios. Please advise

Also I am confused as when to use counting menthod and when to use formulae

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 ?