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A first grade teacher uses ten flash cards, each numbered fr

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A first grade teacher uses ten flash cards, each numbered fr [#permalink] New post 11 Sep 2005, 17:10
00:00
A
B
C
D
E

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Question Stats:

65% (02:46) correct 35% (02:05) wrong based on 62 sessions
A first grade teacher uses ten flash cards, each numbered from 1 to10, to teach her students to order numbers correctly. She has students choose four flash cards randomly, then arrange the cards in ascending order. One day, she removes the cards '2' and '4' from the deck. On that day, how many different correct arrangements of four randomly selected cards are possible?

A. 70
B. 210
C. 336
D. 840
E. 1680
[Reveal] Spoiler: OA

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Last edited by Bunuel on 21 Mar 2014, 00:19, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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 [#permalink] New post 01 Oct 2005, 20:26
This is a combinations problem so you need to use the correct formula:

10-2 cards =8 total cards. all possible permuations of 4 chosen from 8 cards are 8!/4!(8!-4!) = 8!/4!*4! =8*7*6*5*/4*3*2*1* =70

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 [#permalink] New post 03 Oct 2005, 19:06
we need the OA

originally , i calculated
8!/(8-4)! due to mention of the cards being arranged in ascending order.
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 [#permalink] New post 03 Oct 2005, 19:07
UpperTaxBracket wrote:
we need the OA

originally , i calculated
8!/(8-4)! due to mention of the cards being arranged in ascending order.


Ooops... my bad... OA is A.
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 [#permalink] New post 04 Oct 2005, 12:17
i think i'm missing something here, if arrangement/order is a part of this problem why is it considered a combination problem as opposed to a permutation one?
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Re: PR '05 A teacher uses ten flash cards, numbered 1 through [#permalink] New post 20 Mar 2014, 17:17
Yes, I have the same doubt. I really don´t understand the solution.
Book solution:
Make four spots for the cards chosen, and then fill in the number of options for each spot:

8 x 7 x 6 x 5

Then divide by the number of ways to arrange theses four cards: 4 x 3 x 2 x 1
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Re: [#permalink] New post 20 Mar 2014, 20:35
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silentell wrote:
i think i'm missing something here, if arrangement/order is a part of this problem why is it considered a combination problem as opposed to a permutation one?


It's not advisable to tag the questions using its wording. Just because the question uses the word 'arrangement', it doesn't make this an arrangement problem. It is a combination problem and here is why: When you pick the cards, there is only one way in which you can arrange them - the ascending order which will be unique for any 4 cards you pick. Say, I picked up 3, 5, 6, 10. The only way I can arrange them is this: 3, 5, 6, 10. I cannot arrange them as 3, 6, 10, 5 or 6, 5, 3, 10 or any other arrangement that you can have with 4 unique cards. The number of arrangements here is not 4!. Instead it is only 1 since they must be put in ascending order.

So all you have to do is find out the number of ways in which you can pick 4 cards out of 8 unique cards. This will be done in 8C4 = 70 ways.

The book uses the basic counting principle. Put 4 cards in 4 places in 8*7*6*5 ways and since you can arrange them in only one way, divide by the number of arrangements i.e. 4!

When will it be an arrangement problem? "She has students choose 4 cards randomly, then arrange the cards in ANY order."
Now each selection will have 4! distinct arrangements.
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Re:   [#permalink] 20 Mar 2014, 20:35
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A first grade teacher uses ten flash cards, each numbered fr

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