SOLUTION: Roy must elect 3 courses from among 6 courses in group I and 3 courses in group II. If he must take at least 1 of his 3 electives from each group, how many choices does he have? (H

Algebra ->  Permutations -> SOLUTION: Roy must elect 3 courses from among 6 courses in group I and 3 courses in group II. If he must take at least 1 of his 3 electives from each group, how many choices does he have? (H      Log On


   



Question 972737: Roy must elect 3 courses from among 6 courses in group I and 3 courses in group II. If he must take at least 1 of his 3 electives from each group, how many choices does he have? (Hint: First find how many choices he has if he elects only 1 course from group I. Then find how many choices he has if he elects 2 courses from group I. Since he must do one or the other of these, the final answer is the sum of the two answers.)
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Roy must elect 3 courses from among 6 courses in group I and 3 courses in group
II. If he must take at least 1 of his 3 electives from each group, how many
choices does he have?
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%2A2%2F%282%2A1%29=6%2F2 = 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 = 6%2A5%2F%282%2A1%29=30%2F2 = 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.

the sum of the two answers.

18 + 45 = 63 ways total.

Edwin