Question

In how many ways can a person draw 5 cards from a 52 card deck and get:
a. Three aces and a pair
b. Three of a kind and a pair

So I know you have to use the choose, C(_,_), formula but am having a bit of trouble with how you combine the ways of getting the two types required. With choosing 4 aces out of the deck, you just remove 4 cards and for the second part just do C(48,_) I think. I'm confused as to what you do when you have 3 since I don't think you can just remove 3 cards since there are different ways of getting 3 aces(my guess would be 4 ways of getting 3 aces).

Answer 1

Usually you have to split larger tasks into smaller sequential/simultaneous tasks.

Answer 2

Shall I demonstrate how I would do A for you?

Answer 3

@Its_Sonic Listen up

Answer 4

Yes that would be great, that would give me the right idea on how to approach these questions :)

Answer 5

a
i - pick 3 of the 4 aces: \(^4C_3\)
ii - pick 1 other type of pair \(^{12}C_1\)
iii - pick 2 cards out of the four which are the other type of pair: \(^{4}C_2\)

These are all sequential tasks. Sequential tasks are combined by multiplication.\[
^4C_3\times ^{12}C_1\times ^4C_2
\]

Answer 6

Make sense?

Answer 7

Since \(^nC_1=n\), lots of people tend to not write it as a choose. I like being thorough about it though.

Answer 8

So theres 12 types of pairs because there are 13 cards and you can't make a pair of aces since there are 3 chosen already? This makes it clearer for me, my course makes the choose definition very confusing >_< Thanks!

Answer 9

There are many interpretations of chose.

Answer 10

My interpretation is: take items out of a bag, it doesn't matter the order in which items were removed.

Answer 11

But you can use it, for example, to assign a position as well.