Question

In a deck of 52 cards, find how many five-card hands are possible that contain at least three of a kind?

Answer 1

these are one of many problems that i have problems with...

Answer 2

the answer is 59280, how to find it??

Answer 3

please help, im so stumped xD i have a test tomorrow

Answer 4

Any such hand can have at least a three-of-a-kind or at most a four-of-a-kind. Consider the first case. The first three cards have \(13\dbinom43\) ways of showing up in the hand - there are 13 different card values we can get (A, 2, 3, etc), and for any one of these you have 4 possible suits of which you only get 3. There are 49 cards remaining in the deck, or 48 not counting the value in the three-of-a-kind. Of these cards you choose 2, i.e. the other two cards contribute \(\dbinom{50}2\) ways. In total: \(13\dbinom43\dbinom{48}2\).

Now consider the second case. The first four cards have \(13\dbinom44\) ways of showing up. The last card is any one of the 48 remaining cards. In total: \(13\dbinom44\dbinom{48}1\).

Answer 5

Whoops, that \(\dbinom{50}2\) should read \(\dbinom{\color{red}{48}}2\).

Answer 6

uhm, how do i read the notation?

Answer 7

\(\dbinom nk={}_nC_{k}=\dfrac{n!}{k!(n-k)!}\)

Answer 8

ok

Answer 9

I assume you know the other two forms? I prefer the binomial coefficient \(\dbinom nk\) notation myself.

Answer 10

oh so basically, you have to multiply 13?

Answer 11

Yes, the \(\dbinom43\) accounts for the number of ways any ONE three-of-a-kind might show up, so you need to multiply by 13 to account for ALL the possible three-of-a-kind hands.

Answer 12

And finally, of course, you have to add these totals to get the answer, 59280.

Answer 13

ah okay! thank you omg

Answer 14

You're welcome!

Answer 15

do you mind answering a few more?

Answer 16

i've marked like 4 questionis that i was stumped on.

Answer 17

Let's see what we can take care of.

Answer 18

ok thank you so much. im dying from exhaustion..

Answer 19

A test has five true-false problems. Assume all problems are answered.
- If you know there are more true answers than false answers, then in how many different ways can you answer the problem.

Answer 20

Since there are more "trues" than "falses", you can have 3, 4, or all 5 answers be true. Take it case-by-case again.

If only 3 are true, the other 2 must be false. Since these problems are presumably distinct, we're counting the permutations, not combinations. In other words, answering TTTFF is NOT the same as FTTTF, and so on. Of the five questions, three must be true, so you have \({}_5P_3=\dfrac{5!}{(5-3)!}\) ways of having three true and two false answers.

The same reasoning holds for and 5 true answers. You'll end up summing them all,
\[\sum_{k=3}^5 {}_5P_k={}_5P_3+{}_5P_4+{}_5P_5\]

Answer 21

oh ok , so does that mean 16 is not the answer? cause the answer key say it's 16... but the sum is 120

Answer 22

Well the sum above ends up being 300, not 120. If that's not right, then they must be assuming that TTTFF is the same as guessing FTTTF. In that case, you're summing
\[\sum_{k=3}^5 {}_5\color{red}C_k={}_5\color{red}C_3+{}_5\color{red}C_4+{}_5\color{red}C_5\]
Sorry for the mix-up, they tend to happen when the question isn't written clearly.

Answer 23

wow, you get these so quickly... i guess it takes a lot of experience.. );

Answer 24

There's a whole discipline devoted to counting things! And yes, it takes practice, but at some point it all begins to make sense and just clicks.

Answer 25

ok, i have like 3 more quick and simple ones that i dont understand. haha

also,

find the number of diagonals of a regular polygon with 30 sides? (i dont see how this relates)

Answer 26

There's a quick formula for this scenario, but I always find it useful to derive it through example.

Let's start with a regular hexagon (6 sides). From any one vertex, we have 3 possible diagonals we can draw.

We only have 3 since we can't draw a diagonal from a vertex to itself, and we can't draw a diagonal to adjacent vertices.

Answer 27

From the next vertex, we again have 3 possible diagonals:

Answer 28

From the next, we again have 3, but one of these is already drawn, so in fact we only add 2 new diagonals: