ADA test- permutations

[RUmolarman]

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I know this threat has been covered millions of times: but heres the question:
Jill has 6 different books. In how many ways can jill select two different books?

answer- 30


Wouldnt you use n!/r!(n-r!) because order DOES NOT matter?

This answer suggests the use of n!/(n-r)!... 6!/4!= 30. How??

 

[Streetwolf]

I know this threat has been covered millions of times: but heres the question:
Jill has 6 different books. In how many ways can jill select two different books?

answer- 30


Wouldnt you use n!/r!(n-r!) because order DOES NOT matter?

This answer suggests the use of n!/(n-r)!... 6!/4!= 30. How??

Ya it's a stupid question, poorly worded.

I guess if the first book she chose she would read first, then it would matter. Otherwise I can't see why it would be a permutation.

 

[z3u2]

yeah i guess it may seem ambigous and is quite tricky.

but i think it's correct.

here's another example with similar question, it makes a good practice and will help u to understand the original question:

In how many ways can u roll two dice that results with different numbers on each die? U may think rolling 2 and 3 will be same as rolling 3 and 2, but no, order does matter and the answer will also be 30.

explanation:
U have two dice, dice A and dice B.
Suppose first throw: dice A shows 2, dice B shows 3
second throw: dice A shows 3, dice B shows 2
These are two distinctive outcomes, even though both cases have 2 and 3, they are not the same.
If this is hard to understand, try imagining you have one blue dice and one red dice. Clearly, "blue dice with 2/red dice with 3" is different with "red dice with 2/blue dice with 3".

So, back to the original question: How many WAYS can you select two different books? Suppose u have books A, B, C, D, E, F. Selecting book A first then select book B second is one way. Selecting book B first then select book A second is another way. And even though the outcomes are both book A and B, but the WAY to do it is different. So, order does matter. Thus the answer would be 6 * 5 = 30

If, however, the question is changed to "how many combinations can there be when selecting two books out of 6 different books?"
It is asking for the end result, so the process doesn't matter. Selecting 2 first/3 second or 3 first/2second will both result in ONE same combination. Therefore the answer is 6 choose 2 = 15

i hope this helps

 

[Streetwolf]

Yeah it's close but I still think they are different.

For the dice you have two objects that already exist out there as separate items. You have the die on the left and the die on the right (or wherever they are). The one on the left can be any of 6 numbers and the one on the right can be any of 6 numbers.

For the books you don't have 2 objects waiting to be chosen as one of 6 titles... You don't have any blank sheets of paper that you can flip and have magically turn into random reading material. You have 6 books in front of you that you can select.

I think that's where the difference is.

 

[z3u2]

the two dice example was to help visualizing "the way to select"...

we can change the question to with only ONE die, roll first time, and then roll second time.
How many ways can the first and second outcome be different?
How many ways can the two outcomes be different?
How many ways can there be two different outcomes (two different books)?

how about if we have 6 raffle tickets in a box. Each raffle tickets represents a person. We are to draw the raffle tickets twice at random, with replacement.
In how many ways can the two draw to result in the same person? 6 * 1 = 6
In how many ways can the two draw to result in different person? 6 * 5 = 30

What about, we draw the raffle tickets twice, without replacement.
In how many ways can the two draw to result in the same person? 6 * 0 = 0
In how many ways can the two draw to result in different person? 6 * 5 = 30

Back to the book question:
Jill has 6 different books. In how many ways can jill select two different books, with replacement? 6 * 5 = 30
Jill has 6 different books. In how many ways can jill select two different books, without replacement? 6 * 5 = 30
Jill has 6 different books. In how many ways can jill select two different books (with or without replacement)?

With or without replacement, the answer will be the same.
However, if we are now asking for the probability of having two different books, whether there's replacement or not will affect the answer.

And yeah, i think the original question can easily be confused. But i do think it's correct.

 

[RUmolarman]

a