Combination or permutation? : Quant Question Archive [LOCKED]
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# Combination or permutation?

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24 Feb 2007, 18:16
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I would think this is a combination question because order does not matter. As long as 2 books are picked, it doesnt matter which book is picked. Is the answer given by the book wrong or someone please explain why?

QUESTION:
How many different ways can 4 books be arranged 2 at a time?
A. 15
B. 4
C. 24
D. 3
E. 12

The order of arrangement matters in this problem. Therefore, this is a permutation problem. The number of permutations of n objects taken r at a time is:

P(n,r) = n!/(n-r)!
P(4,2) = 4!/(2!)
P(4,2) = 24/2
P(4,2) = 12
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24 Feb 2007, 21:36
I agree to the provided answer.

The question asks "arrange" 2 at a time, so order is important
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26 Feb 2007, 19:46
muvjot wrote:
I would think this is a combination question because order does not matter. As long as 2 books are picked, it doesnt matter which book is picked. Is the answer given by the book wrong or someone please explain why?

QUESTION:
How many different ways can 4 books be arranged 2 at a time?
A. 15
B. 4
C. 24
D. 3
E. 12

The order of arrangement matters in this problem. Therefore, this is a permutation problem. The number of permutations of n objects taken r at a time is:

P(n,r) = n!/(n-r)!
P(4,2) = 4!/(2!)
P(4,2) = 24/2
P(4,2) = 12

How many ways you can select 2 from 4C2 = 6
Each of this selection can be arrange in 2! ways = 2

So it is a combintion and permutation problem.
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17 Mar 2007, 10:43
hopefully my logic is sound, counting the outcomes:

4 outcomes for the first selection * 3 outcomes for the second selection= 12
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17 Mar 2007, 11:45
I dont get what they mean by 2 at a time
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18 Mar 2007, 06:26
"arrangement" always refers to Permutation, because in this case the order becomes relevant. eg in how many ways 5 students can be made to sit in 4 seats - answer is P(5,4)

"selection" always refers to Combination eg in how many ways a committee of 4 students be made from students - answer is C(5,4)
Combination or permutation?   [#permalink] 18 Mar 2007, 06:26
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