Question

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).

Answer 1

6C3+1

Answer 2

it means 6C3 + 1 = 20 + 1 = 21 ?

Answer 3

Could you explain it clearly ? Thank you

Answer 4

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

Answer 5

Obviously you add these up.

Answer 6

So the question is, how to split up 12, 11, 10, ... 6 betwee 2 flavors

Answer 7

If you start with 12 chocolate and 0 blueberry, you can get any other option by changing a chocolate to a blueberry.

Answer 8

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.

Answer 9

So given 'n' cones, we have 'n+1' options to split then between flavors.

Answer 10

So the result is: \[ \Large
\sum_{n=6}^{12} n+1 = \sum_{n=1}^{12} (n+1) - \sum_{n=1}^{5} (n+1)
\]

Answer 11

Ok...i am gonna explain it.......
HERE,THE QUESTIONS SAYS WE MUST HAVE 6 ICECREAM VANILLA FLAVOUR,WHICH IS EQUAL TO ONLY 1 POSSIBLITY....ie=1
And now we can select 6 more ice creams of 3 differnt flavours =6C3
Total possibilties=6C3+1=20+1=21

Answer 12

@Annchelijk Get it?

Answer 13

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