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# Was revising the questions I got from my prep classes and

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Was revising the questions I got from my prep classes and [#permalink]

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24 Apr 2007, 16:01
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Was revising the questions I got from my prep classes and this one came up that doesnt make sense to me....

How many different ways are there to select 2 cards from a standard deck of cards AND obtain 2 jacks?

There are no answer choices, just an OA which I'll post after you've had the kind of fun I had with this one
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24 Apr 2007, 16:57
Okay, I am sure that I am solving it wrong. Anyway,

There are only 4 ways to pick a jack .. then only 3 ways to pick the other one. So, is the answer 12?

Edit: Well my answer seems too simple. Maybe I haven't understood the problem (Darn, that's not a good sign ). riverripper, can you please elaborate on your solution?

Last edited by mNeo on 24 Apr 2007, 17:57, edited 3 times in total.
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24 Apr 2007, 17:12
Two ways to doing this...the easy way and the right way. Since you dont get style points the easiest way on this particular on on the for you may be to write out each option and count them out. Its easy when you only use 4 jacks, if it was a picking out two face cards it would be time consuming this way. If you can't think of the right math do whatever works.

The right way is to say there are 4 jacks, since you need to make a pair and one can't pair with itself you start with N-1. Then continue on N-2 until you reach N-N. Sum up the results (N-1)+(N-2)+(N-3)..+..(N-N) = X

So N=4
4-1 = 3
4-2 = 2
4-3 = 1
4-4 = 0

So for this there are 6 options. If it was 5 then there are 10. 6 would be 15...and so on.

Hope my crazy thinking doesnt confuse you.
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24 Apr 2007, 17:58
up

riverripper, can you please elaborate on your solution? And perhaps explain the question too (As it seems that I misunderstood the question)
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24 Apr 2007, 18:03
hi all,

it says how many different ways are there to pick 2 cards from a deck and end up with two jacks. so basically the number of ways to pick 2 jacks is all that matters because it is a subset of the number of ways to pick 2 cards from 52. And since it is the number of ways, i assume that order does matter, because if you pick jack of spades first, then jack of hearts second, that's a different way compared to the alternative. Then the formula should be:

4P2 4! / (4-2)! = 12
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24 Apr 2007, 19:01
There are jack of hearts, spades, diamonds and something (I can't recall, I don't play cards)

ANyway, there should be 4 jacks in a deck of cards.

The number of ways to pick 2 jacks from these 4 = 4C2.
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24 Apr 2007, 20:10
I think that the way of picking a heart and spade is different from the way of picking a spade and then a heart. Point is, it should be 4P2, not 4C2 in my opinion
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24 Apr 2007, 20:16
mNeo wrote:
I think that the way of picking a heart and spade is different from the way of picking a spade and then a heart. Point is, it should be 4P2, not 4C2 in my opinion

Permutation considers order you pick the card. It doesn't matter if you pick a spade first then a heart or vice versa because you still end up with a jack of spade and a jack of heart. Combination should be used.
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24 Apr 2007, 20:19
I understand that. But the question is "How many different ways are there to select 2 cards"

Isn't the way of choosing heart + spades different from the way of choosing spades + heart? I know that the end result is the same, though.

ps. Look at the post by jayt696969. I think I agree with his opinion
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24 Apr 2007, 20:26
mNeo wrote:
I understand that. But the question is "How many different ways are there to select 2 cards"

Isn't the way of choosing heart + spades different from the way of choosing spades + heart? I know that the end result is the same, though.

ps. Look at the post by jayt696969. I think I agree with his opinion

I don't think I agree with that statement. It's like asking how many ways to pair up jack and jane in a two people team from a pool of four people: jack, jane, john and paul. If your argument is true, then the number of ways to pair up jack and jane is 2 -> (jack,jane) and (jane,jack). We know that's not true because that's really just 1 group. Same for the cards, is there any different in (spade,heart) and (heart,spade)? It's really just 1 group.
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24 Apr 2007, 22:57
It must be 4C2

Choosing Spade Jack+ Club Jack is same as Choosing Club Jack + Spade Jack
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25 Apr 2007, 05:11
ywilfred wrote:
There are jack of hearts, spades, diamonds and something (I can't recall, I don't play cards)

ANyway, there should be 4 jacks in a deck of cards.

The number of ways to pick 2 jacks from these 4 = 4C2.

Great work ywilfred!
That's the OA & OE

I thought it would be 52C2
Obviosly that 52 is just a distraction that only seems to have distracted me
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25 Apr 2007, 05:20
riverripper wrote:
So for this there are 6 options. If it was 5 then there are 10. 6 would be 15...and so on.

Hope my crazy thinking doesnt confuse you.

I don't know how you did it riverripper but your solution works!
I cross-checked your results for 5 Jacks & 6 Jacks using ywilfred's method and your method worked there too!

Easy way and the right way... i got to remember that
25 Apr 2007, 05:20
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