In how many ways can u choose a pizza with three different toppings?
964273
If you number the tpopping from 1 to 8, you can have the following combinations of different toppings (numbered for simplicity):\n1) 1, 2, 3\n2) 1, 2, 4\n3) 1, 2, 5\n4) 1, 2, 6\n5) 1, 2, 7\n6) 1, 2, 8\n7) 1, 3, 4\n8) 1, 3, 5\n9) 1, 3, 6\n10) 1, 3, 7\n11) 1, 3, 8\n12) 1, 4, 5\n13) 1, 4, 6\n14) 1, 4, 7\n15) 1, 4, 8\n16) 1, 5, 6\n17) 1, 5, 7\n18) 1, 5, 8\n19) 1, 6, 7\n20) 1, 6, 8\n21) 1, 7, 8\n22) 2, 3, 4\n23) 2, 3, 5\n24) 2, 3, 6\n25) 2, 3, 7\n26) 2, 3, 8\n27) 2, 4, 5\n28) 2, 4, 6\n29) 2, 4, 7\n30) 2, 4, 8\n31) 2, 5, 6\n32) 2, 5, 7\n33) 2, 5, 8\n34) 2, 6, 7\n35) 2, 6, 8\n36) 2, 7, 8\n37) 3, 4, 5\n38) 3, 4, 6\n39) 3, 4, 7\n40) 3, 4, 8\n41) 3, 5, 6\n42) 3, 5, 7\n43) 3, 5, 8\n44) 3, 6, 7\n45) 3, 6, 8\n46) 3, 7, 8\n47) 4, 5, 6\n48) 4, 5, 7\n49) 4, 5, 8\n50) 4, 6, 7\n51) 4, 6, 8\n52) 4, 7, 8\n53) 5, 6, 7\n54) 5, 6, 8\n55) 5, 7, 8\n56) 6, 7, 8\n\nThis means that there are 56 different topping combinations. Since there are 3 types of pizza, 168 (which is 56 * 3) is the number of different pizzas you can choose from here.\n\nAnswer: 168 different pizza type and topping posibilities.
39238
Dominos. They have that unlimited topping deal. Enjoy
783132
Oh Pizza. \n\nI was just about to say that 14 inches sounds perfect but now I know its a pizza i'm thinking it may be a little on the large side.
60406
Scrambled. Sometimes with ham & cheese in it. Favorite pizza toppings are mushrooms & sausage.
351978
is there a third option?
989012
there are many choices
451690
I agree with the food comment.  If you can get a hold of the standard 8-slice pie, you can try the following:\n\nFirst get two volunteers.  Tell them to divide up the pizza in as many ways as they can think of with the following restriction:\n\n(*) The number of slices the first person takes before the second person has a chance to take one is how they must alternate for the remainder of the game.\n\nA win constitutes both volunteers getting the same number of slices.\n\nThen do the same with four volunteers.\n\nIdeally that'll illustrate that a/b can be interpreted as the number of slices that b people will get when we divide a evenly among them (taking 1 slice at a time) or how many groups of b slices (taking b slices at a time) we can get from a.  \n\nIdeally, you can then link multiplication in there somehow by showing that we have a slices or a/b groups of b slices (from the second interpretation) so that a = (a/b)*b.
894379
I like pepperoni pizza.
293813
i love your avatar.I'm making home made pizza from scratch.Yummy.
876016
The function is A (the area of pizza) = pi(D/2)^2 + pi((30-D)/2)^2, where D is the diameter of one of the pizzas.  The domain is 0<=D<=15, because if D is bigger than 15 then the other pizza's diameter is smaller and the situation is identical to an already exmined one.\n\nThe derivative of the funtion is A' = piD/2 - pi(30 - D)/2.  Setting this equal to 0 means that piD/2 = pi(30 - D)/2. That means that the derivative is 0 when D = 30 - D, or when D is 15.  This is a boundary condition anyway, so there are only two cases to consider:  D = 15 and D = 0.  One of these must be the maximum and the other the minimum.  Plugging the values into the equation (I used Excel for this) you get A(0) = 706.85 and A(15) = 353.43.  So your best bet is to get one infinitesimally small pizza and one 30-inch diameter one.  The pizza parlor's best bet is to sell two 15-inch diameter pizzas.
1001580
twist both with your two hands and suck on the third one
247553
Total no of items\n = 7 forks + 4 spoons + 5 knives \n = 16  \n\nIn the experiment of drawing 3 items from among 16\n\nTotal no. of possible choices \n= no. of ways in which the 3 items can be drawn from the total 16\n==> n = C(16,3)\n==> n = 16 * 15 * 14/3 * 2 * 1\n==> n = 560 \n[On the assumption that the drawing is being made without replacement]\n\nLet \n"A", "B", "C" and "D" be the events of the three items drawn being 3 forks, 2 forks and a knife, 3 spoons and 1 fork, 1 knife and 1 spoon respectively\n\nFor Event "A"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         3           0        0          3\nChoices       C(7,3)      C(5,0)    C(4,0)     C(16,3)\n\nNo. of favorable choices\n= No. of ways in which the 3 forks can be drawn from the total 7\n==> m(a) = C(7,3)\n==> m(a) = 7 * 6 * 5/3 * 2 * 1\n==> m(a) = 35\n\nProbability that the three items drawn are forks = m(a)/n\n==> P(A) = 35/560\n==> P(A) = 1/16\n\n\nFor Event "D"\n\n              Forks      Knives    Spoons     Total\nAvailable       7           5        4          16\nTo draw         1           1        1          3\nChoices       C(7,1)      C(5,1)    C(4,1)     C(16,3)\n\nNo. of favorable choices\n= (No. of ways in which 1 fork can be drawn from the total 7)* (No. of ways in which 1 knife can be drawn from the total 5) * (No. of ways in which 1 spoon can be drawn from the total 4)\n==> m(d) = C(7,1) * C(5,1) * C(4,1)\n==> m(d) = 7 * 5 * 4\n==> m(d) = 140\n\nProbabilty for the 3 items drawn to be 1 fork, 1 kinfe and 1 spoon = m(d)/n\n==> P(D) = 140/560\n==> P(D) = 1/4\n\nYou should be able to solve the two parts in between for Events "B" and "C".\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
564027
At least two of them, or exactly 2 of them?\nOf couse, assuming they choose just one recipe for "supper"\n\nProb. that at least 2 of them choose the same recipe.\n\n1 chooses a recipe:\nprob 2 picks diff. recipe = 7/8\nprob 3 choose diff than 1 and 2 = 6/8\nprob 4 diff than all above = 5/8\nprob 5 diff than all above = 4/8\n\nprob they all choose diff = 7/8 x 6/8 x 5/8 x 4/8 = 0.205\nprob at least 2 choose same = 1 - .205 = 0.795 or about 80%\n\nprob exactly 2.\n\n1 chooses recipe\n2 chooses same = 1/8\n3 chooses diff = 7/8\n4 chooses another diff = 6/8\n5 chooses another diff = 5/8\n\nprob 1 and 2 choose same (rest diff) = 0.0513\nhowever 2 and 3 (or any pair) could choose same (rest diif)\nnumber of possible pairs from 5 is (5)(4)/2 = 10\nSo probability that exactly two are same = 10 x .0513 = 0.513 or about 51%. \n\nnote: prob exactly 2 (51%) is less than at least 2 (80%)
640873
Three sets of three.   That way you can substitute in a letter in any one of the groups if you need more unique identifiers.  \n\nYou can also use each set as a sub category, kind of like a VIN number on a car.
1055081
There are three ways to select one ball: R, B, W\nThere are 3 * 2 * 1 ways to select two balls:  RR, RB, RW, BB, BW, WW\nThere are 4 * 3 * 2 * 1 ways to select four balls\nThere are 5 * 4 * 3 * 2 * 1 ways to select five balls\n\nThis is refered to as 5! ways... 5 * 4 * 3 * 2 * 1\n\nThis continues until you hit the number of ways to select 8 balls, because you can't have sevel red balls.  You'd have 8! - 1\nFor 9 balls, you'd have 9! - 2! (there are 2 ways to have more than 7 red balls)\nFor 10 balls, you'd have 10! - 3! - 1 (3 ways to have more than 7 red balls, 1 way to have more than 9 black balls)\n\nIt proceeds like this.\n\nThe actual odds are calculated differently.
16141
There are 6 combinations\n3,5,7\n3,7,5\n5,3,7\n5,7,3\n7,5,3\n7,3,5
816035
three... i love it... i like to eat it, but eating it off some hot guy is much yummier...