There are no limitations mentioned on the question. Thanks for the help. No I'm not cheating, this is a test review so I just need to see how it's done.A restaurant offers pizza with 3 types of cheese and 7 different toppings. How many different combinations.?
Assuming that you can have only one type of cheese and one type of topping on one pizza, there are 21 combinations.
However, if you can have at least one type of cheese and one type of topping on pizza, there are all in all 49 combinations (3*7=21 one cheese+toppic, 3*7=21 two cheeses+toppic, 1*7=7 three cheeses + toppic)
Things get even more complicated if you can have at least one cheese and at least one toppic on one pizza...7(1 toppic on pizza)+21(two toppics on pizza)+15+10+6+1=60 different combinations of 7 topics, so (3+3+1)*60=420 combinations of flavours.
If you add that is possible cheeses only and topics only on pizza(nobody will eat pizza with nothing), there are 7 + 60 combinations more, so all in all there result is 487.
21A restaurant offers pizza with 3 types of cheese and 7 different toppings. How many different combinations.?
21.
7 times 3 = 21
21 different combinationsA restaurant offers pizza with 3 types of cheese and 7 different toppings. How many different combinations.?
draw a diagram if you need to...
21
7*3=21
7 pizzas for fst cheeze,
7 for 2nd,
7 for third
hence 21
7*3=21
Cheese= A, B, C
Toppings = 1..7
A1, A2, A3...etc
B1, B2, B3...etc
...
7*3=21
BUT - what says that I can't have more than one cheese and more than one topping?
A12 or B7123 etc
A1, A12, A13 etc
That would give 3^7=2187
7x7x7x7x7x7x7x3=2470629 flavor combinations
just 1 combination for me thanks
i will have the lot on mine
7 x 3=21
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