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Shuffle a deck of card of 52 (13 different types, one of each of four suits); there are four aces(hearts, diamonds,spades,clubs) cards and pick two cards at random. observe the sequence of two cards in the order in which they were chosen

Mathematics
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a) how many outcome are in sample space it is just P (52 2 )
b)how many outcomes are in events that two cars are same type but different suits?
the sample space is 52x51 and the outcome is 4x4x3

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Other answers:

please explain
Since we've to observe in sequence, we've to do it separately for the first card and for the second card and then, multiply them. For the first card, the sample space is 52 and there will be 4cards of same type and different suite. You can select one out of them in 4ways. For the second card, as you already picked up a card, you will have 51 sample spaces now and as you have to pick up a card of same type as the first and you now have only 3 such cards, you can select it in 3ways. So,the probability of first card x probability of second card = 4/52 x 3/51 As there are 13 types of cards, this can be done in 13ways depending on the first type you might pick. So, you have to multiply the answer you got with 13 and the final answer is 13 x 4/52 x 3/51 So, the outcome is 13x4x3(not 4x4x3...sorry) and sample space is 52x51
I hope you got it
@UnkleRhaukus above answer doesn't take into account that order does matter
there are only two ways to order two cards
I think we have to use permuatation. 13*(4 P 2)
good point
If you pick up the two cards at the same time, then the answer is 13*(4 C 2) You cannot use permutation here, instead use combination
sample space is 52
My first solution was based on picking the cards separately
what if order actually didn't matter; how would we change it?

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