Can anyone walk me through this?
Leo can order a small, medium, or large yogurt. There are 10 flavors and 5 toppings from which to choose.
How many ways can Leo order a yogurt?
A.
18
B.
35
C.
50
D.
150
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OpenStudy (fibonaccichick666):
OOH! Funzies yes I can!
OpenStudy (fibonaccichick666):
SO let us begin. How many sizes can he choose from?
OpenStudy (anonymous):
3
OpenStudy (anonymous):
What a lame yogurt store they are to ONLY offer us 1 topping!
OpenStudy (fibonaccichick666):
So now, let's say he picks large, how many sizes are left?
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OpenStudy (anonymous):
um.. 2
OpenStudy (fibonaccichick666):
good and now if he got a medium too? How many left?
OpenStudy (anonymous):
1
OpenStudy (anonymous):
How is that at all relevant?
OpenStudy (anonymous):
3 choices for size x 10 choices for flavor x 5 choices for topping
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OpenStudy (anonymous):
3 independent choices...so you just multiply
OpenStudy (anonymous):
10*5 ?? =35
OpenStudy (anonymous):
no...3 x 10 x 5
OpenStudy (anonymous):
oh... = 150
OpenStudy (fibonaccichick666):
Alright, so now, what you just did was break that down into a factorial. And because I was thinking permutations not combinations that is not relevant unfortunately
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OpenStudy (anonymous):
yes..150
OpenStudy (fibonaccichick666):
ie 3!=3*2*1
OpenStudy (anonymous):
Factorials are not involved on this one
OpenStudy (fibonaccichick666):
but anyways, this is not a permutations so oopsie
OpenStudy (anonymous):
so the correct answer is 150.... Hmm... that was sort of easy
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OpenStudy (anonymous):
If you had 3 pants and 5 shirts, how many different outfits could you wear?
3 x 5 = 15
same idea heare
OpenStudy (anonymous):
yep...pretty easy
OpenStudy (anonymous):
ohh ok so were multilying here
OpenStudy (fibonaccichick666):
yup
OpenStudy (anonymous):
yes....if the choices you are making are independent of one another...you just multiply the number of choices you have for each decision
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