An ice cream store sells 30 different flavours of ice cream and it offers a choice of 3 different kinds of cones. In how many ways can we order a dozen two-scoop ice cream cones if any two of them in one order must differ at least by a flavor or by the kinds of cones?
Thanks for your help.
Assuming that strawberry on vanilla is considered the same flavour as vanilla on strawberry, i.e. order of the two scoops doesn't count, then there are 30C2 possible flavours, and 3 cones for a total of n=3(30C2) possible ice creams.
For an order of 12 different ice-creams, there are nC12 possible distinct orders.
To solve this problem, we need to consider the different scenarios and combinations of ordering a dozen two-scoop ice cream cones with certain restrictions.
First, let's calculate the total number of possible combinations without any restrictions. We have 30 different flavors of ice cream and 3 different kinds of cones, so the total number of combinations would be 30 flavors × 3 cones = 90 combinations.
However, we need to consider the restriction that any two cones in one order must differ at least by a flavor or by the kinds of cones. This means that if we choose a specific combination for the first ice cream cone, the second cone must be different either in flavor or in the type of cone, and so on.
Let's break down the problem into two cases:
Case 1: Different flavors, same cones:
If we choose a flavor for the first ice cream cone, there are 30 options. For the second cone, there are only 29 flavors left to choose from, since it must differ in flavor from the first cone. The number of combinations for this case is 30 × 29 = 870.
Case 2: Same flavor, different cones:
If we choose a flavor for the first ice cream cone, there are again 30 options. However, for the second cone, we have 3 choices of cones since it must differ in the type of cone from the first one. The number of combinations for this case is 30 × 3 = 90.
To calculate the total number of combinations, we need to add the combinations from both cases:
Total combinations = Case 1 combinations + Case 2 combinations
= 870 + 90
= 960
Therefore, there are a total of 960 ways to order a dozen two-scoop ice cream cones with the given restrictions.
I hope this explanation helps! Let me know if you need any further clarification.