Case 1:

We find how many choices he has if he elects exactly 1 course from group I. 

A.  6 courses Choose 1 = 6C1 = 6 ways

then for each of those 6 ways to choose 1 from group I, from group II, there are

B.  3 courses Choose 2 = 3C2 =  = 3 ways

to choose 2 courses from group II.

That's 6×3 = 18 ways for case 1.


Case 2:

We find how many choices he has if he elects exactly 2 courses from group I. 

A.  6 courses Choose 2 = 6C2 =  = 15 ways

then for each of those 15 ways to choose 2 from group I, from group II there are

B.  3 courses Choose 1 = 3C1 = 3 ways

to choose 1 courses from group II.

That's 15×3 = 45 ways for case 1.

18 + 45 = 63 ways total.

Edwin

 

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