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.

To solve this problem, we can use the principle of multiplication.

First, let's consider the different choices we have for each aspect of the order:
1. Flavors: We have 30 different flavors of ice cream, and we need to choose 2 scoops of ice cream for each order. Since we can choose any flavor for each scoop, the number of ways to choose flavors is given by the combination (30C2).
2. Kinds of cones: We have 3 different kinds of cones to choose from for each order. Since we need to choose one kind of cone for each order, the number of ways to choose cone types is simply 3.

To find the total number of ways to order a dozen two-scoop ice cream cones, we multiply the number of choices for each aspect together.
Number of ways = (30C2) * 3

Now, let's calculate this expression.

To calculate (30C2), which represents the number of ways to choose 2 flavors out of 30, we need to use the combination formula: nCr = n! / (r!(n-r)!).
In this case, n = 30 and r = 2.

Calculating (30C2):
30! / (2!(30-2)!) = 30 * 29 / (2 * 1) = 435

Now, substitute this value back into the original expression:
Number of ways = 435 * 3 = 1305

Therefore, there are 1,305 different ways to order a dozen two-scoop ice cream cones, given that any two of them in one order must differ at least by a flavor or by the kinds of cones.

To solve this problem, we need to consider the number of options for each part of the order: flavors and cones.

1. Flavors: We can choose any two flavors from the 30 available. To calculate the number of ways to choose 2 flavors, we use the combination formula, denoted as nCr. It is given by:
nCr = n! / (r!(n-r)!)
where n is the total number of options and r is the number of options we choose at a time.

So, for choosing 2 flavors from 30, the calculation would be:
30C2 = 30! / (2!(30-2)!) = 30! / (2! x 28!) = (30 x 29) / 2 = 435

2. Cones: We can choose any two kinds of cones from the 3 available. Using the same combination formula, we have:
3C2 = 3! / (2!(3-2)!) = 3! / (2! x 1!) = 3

3. Combining flavors and cones: Since we need to consider the order in which the flavors and cones are chosen, we multiply the number of options for each part together:
435 (flavors) x 3 (cones) = 1305

Therefore, we can order a dozen (12) two-scoop ice cream cones in 1305 different ways if any two of them in one order must differ at least by a flavor or by the kinds of cones.