The Black Box that is counting (ADV MATERIAL)

[MichelleTran]

So I am taking intro to stats and probability and I am having trouble understanding something that may be quite difficult to explain in counting.

Let me start by telling you what i understand
The choose function i.e (6 C 2) will take six elements and create a SET of 15 TUPLE's, in which order does not matter, it does not over count for example if i had six letters "AB" "BA" it would count one of the permutations but not both, NOTE: we don't know which one it does count, which doesn't matter anyways.

Now we get into an example in probability say
i had 52 cards and they are dealt to 1 person, that person gets 5 cards, So the total possible combination in which (order does not matter) would be a set of 2,598,960 5-TUPLE elements. Which would be in the denominator of the equation.

Now say i want to calculate the probability of getting a flush, My prof pulls a number of out his *** and says its 4*(13 C 5)/(52 C 5). With a very ambiguous explanation on why. So I am doing a little research pattern matching and such. I figure (13 C 5) is 13 cards that have the same suit creating an 5-Tuple of all the possible combination where order does not matter. Then times 4 since there are 52 cards in a deck.

So i figure in order to not under count i must make sure my group size (tuple size) is the same as the group size in the denominator, as well i have to make sure that my set size 13 should be the same size of my denominator set size 52 just via multiplying 13 by 4.
My basic problem is counting and developing the numerator set ,or what i like to call the subset of importance to the solution space i.e denominator, WHAT is the general method for this ? This seems very non intuitive

For instance if i decide to come up with the probability of getting a pair, the answer is
Sheldon Ross says 13*(4 C 2)*(12 C 3)*(4 C 1)*(4 C 1)*(4 C 1)/(52 C 5). in which a , b , c ,d is distinguished and a,a is not distinguishable

When i multiply a choose function with another, ie (5 C 3)*(3 C 2) does it create a set of 5 Tuples in which contains 30 elements?

I know this problem is just an analogue for some general rules in set theory cardinality of some sort, i need to know what the general rules are, so that i can solve "different problems" that roughly have the same analogue.
Does anyone know what i am talking about if not i will try and make it clear, i am really frustrated about this help me please and thanks

 

[mrbohn1]

Well...there are 13 C 5 ways of choosing 5 cards from a given suit. As there are 4 suits, there are 4*(13 C 5) ways of getting a flush.

There are 52 C 5 possible hands altogether, so the probability of getting a flush is just (# ways of getting a flush)/(# possible hands) = 4*(13 C 5)/52 C 5.

I'm not sure if this answers your question(s)or not!

 

[MichelleTran]

my question is not an answer but a theory thanks tho.