Quantatitive Reasoning Help Please

[nt4reall]

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1) Jill has six different books. In how many ways can Jill select two different books .
For this problem , i thought we have to chose 2 books between 6 books . Therefore , i got 6!/(2!)(6-2)!. Then , it should be 15 . However , the solution is 30 ( which is 6!/(6-2)! ) . I really do not know why , and i'm a little confused.

2) 10 is to 2y as 25x is to which of the following ?
a) 5x
b) 5xy
c) 5x/y
d) x/5y
e) 5y/x
THe answer is b . Can anybody explain it ? THanks

 

[ramin123]

The first q is a permutation-- not combination

make another variable-- like Q-- so 10/2y = 25x/ Q

Then solve for Q!!

 

[drzakisadiq]

nt4real,

I hope this will give you an easier way to look at the problem. It says she has 6 different books, but she has to choose 2, so how many combinations can she have?

Draw on a piece of paper two slots. In the first slot, write down the number of books she can choose from. Well, since this is her fist time choosing, she has all 6 to choose from. So put down a 6. Now move to the second slot. Jill now is going to make her second choice. Since she already selected a book from choice 1, she now has 1 less book than she started with. So her options of books to choose from is now down to 5. So put a 5 in the second slot. If she had to pick three books, you would again use the same reasoning, but instead put down a 4 in third slot, because now she only has 4 books left. When you have finished, just multiply the numbers together. In this case, it's only two books, so 6 x 5 = 30. I hope this helps. This is the method that I use rather than trying to memorize some formula because it just confuses me even more. Math is all about reasoning and logic.

 

[drzakisadiq]

Okay, so for the second problem....

The key to problems like these is wording. You have to look for key words. In math, the word IS usually stands for the = sign. And the word OF, will usually indicate multiplication. For instance, if you take a simple percent problem and break down: 20% of 100 is what? Well take 0.20 x 100 = 5. So here, just do the same thing.

10 is to 2y
10 = 2y

25x is to ?
25x = ?

The problem is now a simple proportion where you can cross multiply. Doing that gives 50xy = 10 * ?. Solve for ? and you get 5xy. Hope this helps.

 

[Svart Aske]

nt4real,

I hope this will give you an easier way to look at the problem. It says she has 6 different books, but she has to choose 2, so how many combinations can she have?

Draw on a piece of paper two slots. In the first slot, write down the number of books she can choose from. Well, since this is her fist time choosing, she has all 6 to choose from. So put down a 6. Now move to the second slot. Jill now is going to make her second choice. Since she already selected a book from choice 1, she now has 1 less book than she started with. So her options of books to choose from is now down to 5. So put a 5 in the second slot. If she had to pick three books, you would again use the same reasoning, but instead put down a 4 in third slot, because now she only has 4 books left. When you have finished, just multiply the numbers together. In this case, it's only two books, so 6 x 5 = 30. I hope this helps. This is the method that I use rather than trying to memorize some formula because it just confuses me even more. Math is all about reasoning and logic.

This makes sense. How would one solve a combination question then?

 

[drzakisadiq]

Could you be more specific in what you mean by a combination problem? Like an example maybe if you have one.

 

[Streetwolf]

#1 is poorly worded and has been discussed multiple times here.

 

[Svart Aske]

I think I have it figured out, but my initial confusion was it seemed to me like the OP's question was a combination rather than a permutation (is there anything that suggests it's the latter?) so you apply the formula 6!/(2!x4!) to get 15. A permutation, which you described, is 6!/4! = 30. I guess in order to solve for a combination using your method would just be to divide whatever you got with a permutation by x factorial (x = # of members selected from sample), essentially a roundabout way of knowing the formulae, heh.

 

[nt4reall]

This makes sense. How would one solve a combination question then?

Thanks guys alot . I think it is easy for me to understand the first problem like SvartAske said. The second problem all posts are helpful. However , when will u know the problem is combination or permutation ? Thanks again.

 

[Streetwolf]

This is a bad problem. Ignore it. You will know when it is a combo vs perm. Figure out if order of selection matters or not.

 

[TheGreenWave]

How cliche, streetwolf posting on the thread labeled quantitative reasoning help! haha

What would everyone do for math help if you were not around??

 

[nt4reall]

Thanks drzakisadiq. Your answer is easy to understand.