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vikramjit_01
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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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jayt696969
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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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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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mNeo
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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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mNeo
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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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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mNeo
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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It must be 4C2

Choosing Spade Jack+ Club Jack is same as Choosing Club Jack + Spade Jack
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ywilfred
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 :P
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riverripper

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! :shock:

Easy way and the right way... i got to remember that :wink:



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