Combination Question?

Question

A box contains sweets of 6 different flavours. There are at least 2 sweets of each flavour. A girl selects 3 sweets from the box. Given that these sweets are not all of the same flavour, calculate the number of different ways she can select her 3 sweets.
Please explain your reasoning and give me some tips to help me understand permutation and combination once and for all for my november GCE AddMaths exams :)

Thanks!!

Answer 1

Notation: C(n,k) is the binomial coefficient "n-choose-k", n! / [k!(n-k)!]

She could select 3 sweets of all different flavors. Since there are 6 flavors total, there are C(6,3) ways to do this.

She could select 2 sweets that are the same flavor, and one of another flavor. The number of ways to do this is 6*5, since there are 6 flavors total, and once the 2 sweets of the same flavor are chosen, the 3rd sweet could be any of the other 5 flavors.

She could select 3 sweets that are all the same flavor, but we don't know how many ways there are to do this, since we are only assured that there at least 2 sweets of each flavor. If we knew that there were *exactly* two of each flavor, then we wouldn't have to consider this possibility.

Answer 2

firstly, there is a "mistake" in this question...

unless we know exactly how many total sweets are there in the box, it is not possible to calculate the possibilities...

so we make an assumption here... that there are exactly 2 sweets of each flavour .... making it a total of 12 sweets in 6 flavours....

if the sweets are named, say, x1, x2, ...,x12 then we can select any combination of 3 in 12C3 ways
(mCn here means: m!/(n!(m-n)!)
ie. total possibilities = 12*11*10/3*2 = 220 ways

the number of possibilities of selecting 3 different flavours is calculated as follows:

the first sweet can be choosen in any way: 12 ways
the second sweet can be chosen in 10 ways (disallowing 1 in 11)
the third sweet can be chosen in 8 ways (disallowing 2 in 10)

so the total possibilities = 12*10*8 / 3*2 = 160 (the '3*2' is to eliminate repeated entries such 'abc' 'acb' 'bca'...)

note: the intricacies of the topic of "permutations and combinations" cannot be explained in one paragraph... one needs to read the text many times and practice, practice, practice.... !