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Randolph has a deck of 12 playing cards made up of only 2

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Joined: 11 May 2014
Posts: 17
Re: Randolph has a deck of 12 playing cards made up of only 2  [#permalink]

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New post 24 Oct 2014, 15:10
voodoochild wrote:
Bunuel,
Thanks for finding a similar question. However, I am not sure how to calculate the number of ways to obtain "a pair" using PErmutations. The question that you have posted is different - it asks us to calculate probability. This one is different because it asks us to calculate "combinations." I see that the author has just modified the same question. The source of this question is 700-800 series by MGMAT.....

Please help me :( I am really struggling...thanks



Hi, I found a super easy way to solve it.

1. First, find out the combination of 2 pairs. (1122, 1133, so on) This is simply 5*6=30
2. Second, will find out 1 pair combinations. Assume 1 and 1 are chosen along with 2 another cards. (1,1,x,x)
3. The last 2 cards are chosen from remaining 10 cards. Which means 10!/(2!*8!)
4. Together: 1*45=45
5. There are another 5 pairs and every pair have above number of combinations. So 6*45=270 (11xx, 22xx, 33xx, 44xx, 55xx, and 66xx)
6. BUT think that the combination of remaining 2 cards can be pair too. We already knew it is 30.
7. In order to get rid of this overlap, we subtract. 270-30=240.

I might be wrong. But I still feel quite confident. :roll:
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Re: Randolph has a deck of 12 playing cards made up of only 2  [#permalink]

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New post 24 Oct 2014, 15:11
voodoochild wrote:
Bunuel,
Thanks for finding a similar question. However, I am not sure how to calculate the number of ways to obtain "a pair" using PErmutations. The question that you have posted is different - it asks us to calculate probability. This one is different because it asks us to calculate "combinations." I see that the author has just modified the same question. The source of this question is 700-800 series by MGMAT.....

Please help me :( I am really struggling...thanks



Hi, I found a super easy way to solve it.

1. First, find out the combination of 2 pairs. (1122, 1133, so on) This is simply 5*6=30
2. Second, will find out 1 pair combinations. Assume 1 and 1 are chosen along with 2 another cards. (1,1,x,x)
3. The last 2 cards are chosen from remaining 10 cards. Which means 10!/(2!*8!)
4. Together: 1*45=45
5. There are another 5 pairs and every pair have above number of combinations. So 6*45=270 (11xx, 22xx, 33xx, 44xx, 55xx, and 66xx)
6. BUT think that the combination of remaining 2 cards can be pair too. We already knew it is 30.
7. In order to get rid of this overlap, we subtract. 270-30=240.

I might be wrong. But I still feel quite confident. :roll:
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Re: Randolph has a deck of 12 playing cards made up of only 2  [#permalink]

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New post 26 Oct 2014, 20:49
usre123 wrote:
Hello,
I'm stuck here:

I did 12 * 1*10*8
Number of ways of picking any card out of 12 (first blank), then picking it's pair for the second blank (that's one). Then picking any card from the ten remaining, and then picking any other card making sure that it's not the 3rd card's pair.

I got 960, and I understand that in this method position is imp, whereas in the question it is not. I'm wondering how do I factor this out?
Thanks!


Could someone please help with this? thanks
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Re: Randolph has a deck of 12 playing cards made up of only 2  [#permalink]

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New post 21 Sep 2015, 08:33
usre123 wrote:
usre123 wrote:
Hello,
I'm stuck here:

I did 12 * 1*10*8
Number of ways of picking any card out of 12 (first blank), then picking it's pair for the second blank (that's one). Then picking any card from the ten remaining, and then picking any other card making sure that it's not the 3rd card's pair.

I got 960, and I understand that in this method position is imp, whereas in the question it is not. I'm wondering how do I factor this out?
Thanks!


Could someone please help with this? thanks


Assuming the question means "only one pair" (not a valid assumption though)

You have ordered the cards. Note that you need to select two kinds of cards - one a pair and another a set of two cards which are different.
You can select the pair in 6 ways (pair of 1 or pair of 2 or pair of 3 etc). Another way of saying this is: select the first card of the pair in 12 ways and then the second card in 1 way only. But since there is no first-second arrangement, divide 12 by 2! to get 6.
You select the dissimilar cards in 10*8 ways - 10 ways for the first one and 8 ways for the second one. Again, there is no first-second so you divide this by 2 to get 10*8/2 = 40 ways.

Total you get 6*40 = 240 ways
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Re: Randolph has a deck of 12 playing cards made up of only 2  [#permalink]

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New post 28 Nov 2018, 02:50
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Re: Randolph has a deck of 12 playing cards made up of only 2 &nbs [#permalink] 28 Nov 2018, 02:50

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