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anonymous
 one year ago
In a deck of 52 cards, find how many fivecard hands are possible that contain at least three of a kind?
anonymous
 one year ago
In a deck of 52 cards, find how many fivecard hands are possible that contain at least three of a kind?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0these are one of many problems that i have problems with...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the answer is 59280, how to find it??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please help, im so stumped xD i have a test tomorrow

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Any such hand can have at least a threeofakind or at most a fourofakind. 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 threeofakind. 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
 one year ago
Best ResponseYou've already chosen the best response.0Whoops, that \(\dbinom{50}2\) should read \(\dbinom{\color{red}{48}}2\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0uhm, how do i read the notation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\dbinom nk={}_nC_{k}=\dfrac{n!}{k!(nk)!}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I assume you know the other two forms? I prefer the binomial coefficient \(\dbinom nk\) notation myself.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh so basically, you have to multiply 13?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, the \(\dbinom43\) accounts for the number of ways any ONE threeofakind might show up, so you need to multiply by 13 to account for ALL the possible threeofakind hands.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And finally, of course, you have to add these totals to get the answer, 59280.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ah okay! thank you omg

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you mind answering a few more?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i've marked like 4 questionis that i was stumped on.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's see what we can take care of.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you so much. im dying from exhaustion..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A test has five truefalse 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
 one year ago
Best ResponseYou've already chosen the best response.0Since there are more "trues" than "falses", you can have 3, 4, or all 5 answers be true. Take it casebycase 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 \({}_5P_3=\dfrac{5!}{(53)!}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0oh ok , so does that mean 16 is not the answer? cause the answer key say it's 16... but the sum is 120

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well 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 mixup, they tend to happen when the question isn't written clearly.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wow, you get these so quickly... i guess it takes a lot of experience.. );

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0There'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
 one year ago
Best ResponseYou've already chosen the best response.0ok, 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
 one year ago
Best ResponseYou've already chosen the best response.0There'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
 one year ago
Best ResponseYou've already chosen the best response.0From the next vertex, we again have 3 possible diagonals: dw:1433380581390:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0From 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
 one year ago
Best ResponseYou've already chosen the best response.0Continuing this pattern, we end up with this: dw:1433380652284:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you try this for any general regular polygon with \(n\) sides, you'll get the same pattern: \[\{n3,n3,n4,n5,\ldots,2,1,0,0\}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let's derive the formula. Let's call the total number of diagonals \(S\), then \[S=(n3)+(n3)+(n4)+(n5)+\cdots+2+1+0+0\] Let's rearrange this sum: \[S=0+0+1+2+\cdots+(n5)+(n4)+(n3)+(n3)\] If we add corresponding terms (first to first, second to second, and so on) we get \[2S=((n3)+0)+((n3)+0)+((n4)+1)+\cdots+(0+(n3))\] Simplifying, we end up with \(n3\) showing up \(n\) times (since there are \(n\) sides in the polygon) on the right hand side: \[2S=n(n3)\] Dividing by 2, we get the formula: \[S=\frac{n(n3)}{2}\] 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
 one year ago
Best ResponseYou've already chosen the best response.0O__O my teacher never taught us this!! oh god. but now i know what to do then!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0two 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
 one year ago
Best ResponseYou've already chosen the best response.0Only 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
 one year ago
Best ResponseYou've already chosen the best response.0oh wait, i did that before... i think i calculated it wrong ... ahh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0what do i do when the number of elements (n) is less than the number of spaces (r)?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Perhaps 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 \({}_5P_3=\dfrac{5!}{(53)!}\), but I'm not entirely sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's how i thought about it at first!! but it was wrong ):... maybe there was a mistake to the problem >..

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0also, can you clarify stuff for me on circular permutations?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.0and if there was not specified seat that's special i would just do (n1)! and if we were talking about bracelets/charms if there is a clasp: i would use n!/2 no clasp then i use (n1)!/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure I follow. \(\dfrac{n!}{n}=(n1)!\).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0and i get 120 which is the same as 5!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It 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
 one year ago
Best ResponseYou've already chosen the best response.0but dont i have to follow the formula for circular permutation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not familiar with the formulas that you learned, I'm afraid...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok. but thanks so much. you saved me!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You'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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