It depends on whether you count chocolate on top and strawberry on bottom as different than strawberry on top and chocolate on bottom.
If you want those to be the same, you use 31 choose 2, which is the number of combinations of two items you can have with 31 available items. To calculate this, you use 31! /2!(31-2)! =31x15
If you want those to be different, you'll use permutations instead. This would be 31!/(31-2)! = 31x30
See this website for more information: mathforum org
If you have to have two different single scoop flavors to have a double scoop cone, the answer is (40*39)/2 = 780. (Divide by two because chocolate over vanilla is the same as vanilla over chocolate.) If you allow double scoops of the same flavor, add 40 to the total.
The answer is 992. This assumes that, for example, chocolate and vanilla is the same as vanilla and chocolate, so you have to remove those duplicates (this is why its not 31 squared). The formula is n*([n+1]/2) to get the ice cream combos, then multiply by 2 for the two cone choices. Note: n=31.
This really isn't a food question, but more of a finite math question. The answer is 8 combinations. There are two choices of three things, so 2^3=8. This is a relatively simple problem and it is easy to list all the choices, then count them.
You can make 10 triple deckers and one left to drop to your dog. A technical answer for a math test would be 101/3 three scooper's.
But, presuming you mean that you have more than one scoop of each flavor available:
If more than one scoop on the cone can be of the same flavor, the answer is the product of 31 x 31 x 31, or 29,791 combos.
If each scoop has to be different, then it would be 31x30x29, or 26,970 triple deckers.
Quite a bellyful, either way.
The first can be any one of 31 flavors. For each of those . . .
The second can be any one of 30 flavors.
The total number of ways of selecting 2 flavors is (31 x 30) = 930 ways.
If you stack them on a cone, it's obvious which one you scooped first and which
one was second, so there are 930 different 2-flavor cones.
But if you put them in a dish, then nobody can tell which one you scooped first,
so any combination of flavors can go into the dish in 2 ways.
Although there are 930 different ways to choose two flavors, you can only make
half that many = 465 different two-flavor dishes.
If you take account of which flavour is on top, then 31*30 = 930.
If they are considered the same, then 930/2 = 465.
Ffg
just multiply it so 6x3 is 18 and 18 flavors is your answer!
There are 18 ways: you can choose the ice cream flavour in 6 ways and with each of these ways you can choose a topping in 3 ways.
I think...................21
I think...................21
If there can only be one flavor of ice cream and one topping then 31 times 25 then 775.
In how many ways can fourfour women be selected from the eighteight ​women
Different flavors of ice cream are flavored in different ways. A natural mint chocolate chip ice cream might have mint leaves in it. Mint flavor might also come from mint extract or artificial mint flavors.
10 x 9 x 8 = 720 The first person can choose ten different flavors. This leaves nine choices for the second and then eight for the third, as none of them choose the same type. We multiply these to attain the total number of combinations. (If the match between person and ice cream is unimportant - if we are just concerned about the flavors themselves, regardless of how they are doled out - there are only 720/3! = 120 ways. This is because a selection of flavors A, B and C can be handed to three people in 3! = 6 different ways, which we are not interested in as unique results.)I think its 0 (trick question)
They can be selected in 756 ways.
Well there are six flavors and only two toppings so the easy way to figure it out is six times two which is twelve or you could work it out:vanillanutsgummiebearschocolatenutsgummiebearsstrawberrynutsgummiebearscookies and creamnutsgummiebearspecannutsgummiebearsmint chocolate chipnutsgummiebearsnow just count one ice cream and one topping until they all are used so its twelve :)
If there are 58 defective circuit boards, two can be selected in 58*57/2 = 1653 ways.
18x17= 306 ways