Answer:
10 different groups of three CDs are possible
Step-by-step explanation:
Total CDs are five and 3 has to be chosen where order doesn't matter.
In these kind of scenarios, combinations are used for calculating different ways in which the CDs can be given.
So,
Number of ways in which groups of CDs are possible = āµCā
[tex]= \frac{n!}{r!(n-r)!} \\= \frac{5!}{3!(5-3)!}\\=\frac{5!}{3!2!}\\=\frac{5*4*3!}{3!2!}\\=\frac{5*4}{2}\\=\frac{20}{2}\\=10[/tex]
Therefore, 10 different groups of three CDs are possible ..
