Mouth watering ice creams and mind watering problems-3

We can choose $3$ of the flavors in $\dbinom{6}{3} = 20$ ways.

For each of these $20$ combinations, since we must buy at least one of each of the $3$ flavors, there are $3$ choices for the remaining ice cream we intend to buy. This gives us a total of $20 \times 3 = \boxed{60}$ different ice cream combinations.

Select 3 ice creams from 6 in 6C3 ways and chose one in ice cream from selected ones in 3 ways,so total ways are =>6C3.3=20*3=60 Where "C" represents combinatorial.

Since only 1 ice cream can be of common flavor , it can be selected in $\dbinom{6}{1} = 6$ ways . Not to select the remaining ice creams we can go for $\dbinom{5}{1} \times \dbinom{4}{1} = 20$ . but here we did a blunder that if ice cream A is selected in first part and B in second then it can be that B is selected in first and A in second. So we have to divide it by 2. Our final answer becomes.

$\frac{ \dbinom{5}{1} \times \dbinom{4}{1} \times \dbinom{6}{1}}{2} = \boxed{60}$