anonymous
  • anonymous
In a deck of 52 cards, find how many five-card hands are possible that contain at least three of a kind?
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
these are one of many problems that i have problems with...
anonymous
  • anonymous
the answer is 59280, how to find it??
anonymous
  • anonymous
please help, im so stumped xD i have a test tomorrow

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anonymous
  • anonymous
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: \(13\dbinom44\dbinom{48}1\).
anonymous
  • anonymous
Whoops, that \(\dbinom{50}2\) should read \(\dbinom{\color{red}{48}}2\).
anonymous
  • anonymous
uhm, how do i read the notation?
anonymous
  • anonymous
\(\dbinom nk={}_nC_{k}=\dfrac{n!}{k!(n-k)!}\)
anonymous
  • anonymous
ok
anonymous
  • anonymous
I assume you know the other two forms? I prefer the binomial coefficient \(\dbinom nk\) notation myself.
anonymous
  • anonymous
oh so basically, you have to multiply 13?
anonymous
  • anonymous
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.
anonymous
  • anonymous
And finally, of course, you have to add these totals to get the answer, 59280.
anonymous
  • anonymous
ah okay! thank you omg
anonymous
  • anonymous
You're welcome!
anonymous
  • anonymous
do you mind answering a few more?
anonymous
  • anonymous
i've marked like 4 questionis that i was stumped on.
anonymous
  • anonymous
Let's see what we can take care of.
anonymous
  • anonymous
ok thank you so much. im dying from exhaustion..
anonymous
  • anonymous
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.
anonymous
  • anonymous
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, \[\sum_{k=3}^5 {}_5P_k={}_5P_3+{}_5P_4+{}_5P_5\]
anonymous
  • anonymous
oh ok , so does that mean 16 is not the answer? cause the answer key say it's 16... but the sum is 120
anonymous
  • anonymous
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.
anonymous
  • anonymous
wow, you get these so quickly... i guess it takes a lot of experience.. );
anonymous
  • anonymous
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.
anonymous
  • anonymous
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)
anonymous
  • anonymous
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. |dw:1433380461631:dw| 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.
anonymous
  • anonymous
From the next vertex, we again have 3 possible diagonals: |dw:1433380581390:dw|
anonymous
  • anonymous
From the next, we again have 3, but one of these is already drawn, so in fact we only add 2 new diagonals: |dw:1433380603875:dw|
anonymous
  • anonymous
Continuing this pattern, we end up with this: |dw:1433380652284:dw|
anonymous
  • anonymous
If you try this for any general regular polygon with \(n\) sides, you'll get the same pattern: \[\{n-3,n-3,n-4,n-5,\ldots,2,1,0,0\}\]
anonymous
  • anonymous
Let's derive the formula. Let's call the total number of diagonals \(S\), then \[S=(n-3)+(n-3)+(n-4)+(n-5)+\cdots+2+1+0+0\] Let's rearrange this sum: \[S=0+0+1+2+\cdots+(n-5)+(n-4)+(n-3)+(n-3)\] If we add corresponding terms (first to first, second to second, and so on) we get \[2S=((n-3)+0)+((n-3)+0)+((n-4)+1)+\cdots+(0+(n-3))\] Simplifying, we end up with \(n-3\) showing up \(n\) times (since there are \(n\) sides in the polygon) on the right hand side: \[2S=n(n-3)\] Dividing by 2, we get the formula: \[S=\frac{n(n-3)}{2}\] To check that this hold for our \(n=6\) example, we get \(S=\dfrac{6\times3}{2}=9\). Counting the diagonals in the picture, we get the same number. Plug in \(n=30\) and you'll get your answer.
anonymous
  • anonymous
O__O my teacher never taught us this!! oh god. but now i know what to do then!
anonymous
  • anonymous
two more i promise ): Five boys and five girls were nomnated for homecoming king and queen. How many ways can a boy be king, a girl be queen and her court of two girls be chosen?
anonymous
  • anonymous
Only one boy can be a king: \(\dbinom51\). Only one girl can be a queen: \(\dbinom51\). There are now four girls left. The court of girls can only consist of two girls: \(\dbinom42\). Multiply.
anonymous
  • anonymous
oh wait, i did that before... i think i calculated it wrong ... ahh
anonymous
  • anonymous
okay, one last more. i just recently learned about permutation circular. am im not sure how would would do this question. find the number of ways in which three peupils can be seated - in a row containing 5 seats and also - in a circle containing 5 seats
anonymous
  • anonymous
what do i do when the number of elements (n) is less than the number of spaces (r)?
anonymous
  • anonymous
Perhaps this isn't the intended way of thinking about this, but suppose we represent a filled chair by the number 1 and an empty chair by 0, so a string of 1s and 0s represents the seating arrangements. For example, if we have three students sitting with an empty chair between them, we have \(10101\); if the students sit together we can have \(11100\) or \(01110\) or \(00111\), etc. Of these 5 positions, we're filling 3: \(\dbinom53\). But this count doesn't consider the students to be distinct, so to account for this we multiply by \(3!\). This is because the 1s can be arranged from left to right as \(\color{red}1\color{blue}11\) or \(\color{blue}11\color{red}1\), etc. There are \(3!\) ways of arranging them this way. So I'm betting the answer is \({}_5P_3=\dfrac{5!}{(5-3)!}\), but I'm not entirely sure.
anonymous
  • anonymous
that's how i thought about it at first!! but it was wrong ):... maybe there was a mistake to the problem >..
anonymous
  • anonymous
also, can you clarify stuff for me on circular permutations?
anonymous
  • anonymous
so if the problem was about a table. -when there is a specified designation or seat, (clasp) then i use the n!/ n formula right?
anonymous
  • anonymous
and if there was not specified seat that's special i would just do (n-1)! and if we were talking about bracelets/charms -if there is a clasp: i would use n!/2 no clasp then i use (n-1)!/2
anonymous
  • anonymous
I'm not sure I follow. \(\dfrac{n!}{n}=(n-1)!\).
anonymous
  • anonymous
for this problem In how many ways can 6 people stand in a ring around the player who is "it" for example. there is a clasp so i would use n!/n?
anonymous
  • anonymous
and i get 120 which is the same as 5!
anonymous
  • anonymous
It seems to me that there are \(6!\) ways. There are 7 people in total, and 6 of them are standing around 1. We're only concerned with the number of ways the 6 people can arrange themselves.
anonymous
  • anonymous
but dont i have to follow the formula for circular permutation?
anonymous
  • anonymous
I'm not familiar with the formulas that you learned, I'm afraid...
anonymous
  • anonymous
ok. but thanks so much. you saved me!!
anonymous
  • anonymous
You're welcome. Feel free to post a new question about circular permutations, I'm sure someone else will be able to help you out.

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