Answer:
120 combinations of CDs are possible
Step-by-step explanation:
The total combinations could be
[tex]10C_3[/tex]
[tex]\frac{10*9*8*7!}{7!*3!}[/tex]
On further expanding, we get -
[tex]\frac{10*9*8}{3!} \\\frac{10*9*8}{3*2*1} \\= 10*3*4\\= 120[/tex]
Hence, 120 combinations of CDs are possible
