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An ice cream parlor sell ice creams with three different fl‡avors: blue- berry, chocolates, and vanilla. How many ways are there to choose a dozen ice creams with at most six vanilla ice creams? (Note: one dozen consists of twelve ice creams).

Discrete Math
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it means 6C3 + 1 = 20 + 1 = 21 ?
Could you explain it clearly ? Thank you

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Other answers:

What I would do is this: 0 vanilla, 12 of chocolate or blueberry 1 vanilla, 11 of chocolate or blueberry 2 vanilla, 10 of chocolate or blueberry ... 6 vanilla, 6 of chocolate or blueberry
Obviously you add these up.
So the question is, how to split up 12, 11, 10, ... 6 betwee 2 flavors
If you start with 12 chocolate and 0 blueberry, you can get any other option by changing a chocolate to a blueberry.
So you have (12, 0), (11, 1), (10, 2), ... , (0, 12). As you can see, the number of options is 0 to 12... i.e. 13.
So given 'n' cones, we have 'n+1' options to split then between flavors.
So the result is: \[ \Large \sum_{n=6}^{12} n+1 = \sum_{n=1}^{12} (n+1) - \sum_{n=1}^{5} (n+1) \]
Ok...i am gonna explain it....... HERE,THE QUESTIONS SAYS WE MUST HAVE 6 ICECREAM VANILLA FLAVOUR,WHICH IS EQUAL TO ONLY 1 And now we can select 6 more ice creams of 3 differnt flavours =6C3 Total possibilties=6C3+1=20+1=21
@Annchelijk Get it?
The question requires AT MOST 6 vanillas. So Ways to choose AT LEAST 7 vanillas=5C3 Ways to choose 12 ice creams = 12C3 Ways to choose AT MOST 6 vanillas = 12C3-5C3

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