18
32 combinations. 4 of these will have no toppings, or all three toppings, 12 will have one topping and another 12 will have 2 toppings.
200 different combinations. :)
three
2*2*2*2 = 16, counting one with no toppings.
Call the toppings topping A, B, C, and D. Each pizza needs 2 toppings, so first think about topping A. There can be AB, AC, and AD. A is finished. Now think about B. AB=BA so you can't use BA. The remaining are BC and BD. Then think about C. You can't do CA or CB because those are already used with AC and BC. The only one left is CD. With D, everything is already used, so let's stack up all the answers. AB AC AD BC BD CD Count em all up and you get 6 different combinations. Chart it up however you wish.
32 combinations. 4 of these will have no toppings, or all three toppings, 12 will have one topping and another 12 will have 2 toppings.
200 different combinations. :)
16 i think
well, you can to topping 1&2, topping 2&3, topping 1&3, topping 1, 2 and 3, and you can also do all three toppings. so that's seven different types for one size pizza, and you can have all combinations in four sizes. that makes a total of 28 different pizza combinations.
16 1 combination of all 4 4 combinations of 3 6 combinations of 2 4 combinations of 1 1 combination of 0
My understanding is that the correct answer is 2,147,483,648 different combinations, but I don't know how to reach this answer. Can someone explain it to me?
Well, honey, if you've got 5 toppings to choose from, you can make a total of 31 different combinations on your pizza. It's simple math - you take 2 to the power of 5 (2^5), which equals 32, then subtract 1 because you can't have a pizza with no toppings (unless you're a monster). So, go wild and mix and match those toppings to create your perfect pizza masterpiece!
you can make 27 triple dip cone combinations from three flavours 3 flavors for the first dip x 3 flavors for second x 3 flavors for third dip = 27 combinations
book
If you have n flavors available, you can make n(n-1) different permutations (each flavor can be on bottom or top and its considered different) or n(n-1)/2 different combinations (unique pairs of flavors ignoring order).I think the answer you want is n(n-1)/2.
13
three