One problem two solutions

[Steven G]

Ms. Lim is old-school and still owns CDs. She has 4 Jazz CDs, 3 pop CDs, and 5 classical CDs. She wishes to select 3 CDs to take on her road trip, but she wants to have at least one Jazz CD. How many different selections could she make?

1st method: 1st pick one Jazz CD AND THEN choose the remaining 2 CDs from the remaining 11. This gives 4*₁₁C₂=220
Now this result equals ₁₂C₃!! Why?

2nd method: Choose 3 from 12 and then subtract off the cases where there are no Jazz CDs. This gives ₁₂C₃ - ₈C₃ = 164

Now I know that (4 choose 1)*(8 choose 2) + (4 choose 2)*(8 choose 1) + ( 4 choose 3)*( 8 choose 0)=164 must be the correct answer.

Why is the 1st method wrong???

 

[Dr.Peterson]

In the first method, you might pick J1 and include J2 among the others, or pick J2 and include J1 among the others; so you will count any selection including both J1 and J2 (at least) twice. You're overcounting.