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Mathematics 17 Online
OpenStudy (loser66):

The "eating club" is hosting a make-your-own sundae at which the following are provided: Flavors: Chocolate, Cookie, Strawberry, Vanilla Toppings: Caramel, Hot fudge, Marshmallow, M&M, Nuts, Strawberries. a) How many sundaes are possible using one flavor and 3 different toppings? b) How many sundaes are possible using one flavor and from 0 to 6 toppings? (I don't get this part) c) How many different combinations of flavors of three scoops of flavors are possible if it is permissible to make all 3 scoops the same flavors? Please, explain.

OpenStudy (loser66):

for part a) I got 4*6C3= 4*20 =80, is it right?

OpenStudy (loser66):

@kirbykirby

ganeshie8 (ganeshie8):

part a is right

ganeshie8 (ganeshie8):

4C1*6C3

OpenStudy (loser66):

Why do we have to use 4C1 ? is it not that "using 1 flavor" exactly 4 ways?

ganeshie8 (ganeshie8):

4C1 : number of ways of choosing 1 flavor from the available 4 flavors

ganeshie8 (ganeshie8):

4C1 = 4

ganeshie8 (ganeshie8):

4C1 tells explicitly that you're picking 1 flavor from available 4 flavors 4 doesn't tell you anything

OpenStudy (loser66):

got you. Explain me part b, please

ganeshie8 (ganeshie8):

you can choose one flavor in : 4C1 ways

ganeshie8 (ganeshie8):

next, for toppings, you're free to have any number you want between 0->6

ganeshie8 (ganeshie8):

you can have either 0 toppings or 1 toppings or 2 toppings ... or 6 toppings right ?

OpenStudy (loser66):

Yes

ganeshie8 (ganeshie8):

lets assume you're taking all different toppings

ganeshie8 (ganeshie8):

then, number of ways of choosing 0 toppings from available 6 toppings = 6C0

ganeshie8 (ganeshie8):

number of ways of choosing 1 toppings from available 6 toppings = 6C1

ganeshie8 (ganeshie8):

number of ways of choosing 2 toppings from available 6 toppings = 6C2

ganeshie8 (ganeshie8):

...

ganeshie8 (ganeshie8):

number of ways of choosing 6 toppings from available 6 toppings = 6C6

ganeshie8 (ganeshie8):

So total number of ways of choosing 0->6 different toppings = 6C0 + 6C1 + ... + 6C6

ganeshie8 (ganeshie8):

adding them all should give you 2^6 = 64

ganeshie8 (ganeshie8):

Overall, part b answer would be : 4C1 * 64

OpenStudy (loser66):

Gooooooooooooooot it. :)

ganeshie8 (ganeshie8):

Note that we have assumed you're choosing DIFFERENT toppings.

ganeshie8 (ganeshie8):

you're not allowed to pick more than 1 of any of the same toppings

ganeshie8 (ganeshie8):

for example, you're not allowed to pick 2 Caramel toppings even if you like Caramel so much

OpenStudy (loser66):

Yes,

OpenStudy (loser66):

Part c is just about the flavor? so that we ignore the toppings part, right?

OpenStudy (loser66):

And although they say "the combination" but "it is permissible to make all 3 scoops the same flavor" means permutation, right?

ganeshie8 (ganeshie8):

not exactly

OpenStudy (loser66):

:(

ganeshie8 (ganeshie8):

because order doesn't matter when you eat

ganeshie8 (ganeshie8):

`Chocolate, Cookie, Strawberry` is same as `Cookie, Strawberry, Chocolate`

OpenStudy (loser66):

Yes,

OpenStudy (loser66):

I have to go. Please leave your guidance here. I'll be back to take. Thanks in advance. :)

ganeshie8 (ganeshie8):

familiar with stars and bars ?

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