4
2007-12-13 14:19:31
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answer #1
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answered by skippa 5
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There are a total of 4 fives and 4 sixes in each 52-card deck.
2007-12-13 14:35:34
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answer #2
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answered by jean 2
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4
2007-12-13 14:30:55
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answer #3
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answered by ♡♥♡TIA♥♡♥ 2
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Is this a trick question or are you serious, must be 36 of one and 25 of the other ain't there. But how many decks are you counting???
OK if there were 4 people sitting at one table and 4 at another
(4+4=8 altogether right)
and they all buy a round of drinks each costing $1 a drink. It would equal
4 x 4 = 16
4 x 4 = 16
16 + 16 = 32
OK 3 people sit at another table and 5 at another. That is again 8 people altogether. The 3 people buy a round each and the 5 people buy around each
3 x 3 = 9
5 x 5 = 25
9 + 25 = 34
So how come these paid $2 more?
PS some cards have same number printed on them more than once don't they.
2007-12-13 14:22:47
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answer #4
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answered by Anonymous
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4
2007-12-13 14:22:37
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answer #5
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answered by mickkooz 4
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there are 4 of any card in a 52 card deck (4 fives and 4 sixes)
2007-12-13 14:20:51
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answer #6
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answered by Anonymous
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LOL.
In 52 DECKS there would be 208 of each.
4 Suits in each Deck.
4 X 52 = 208
208 + 208 = 416
2007-12-13 14:22:13
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answer #7
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answered by ? 6
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Yes, one five and one six for each of the four suits. Do the math.
Suits = four, hearts, diamonds, spades, clubs.
therefore, 4 of each.
13 cards in each suit. Ace through 10, plus king and queen.
13 x 4 = 52
2007-12-13 14:23:21
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answer #8
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answered by oklatom 7
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This questions pretty much already been answered.
But I would just like to say, "Awe, you silly!"
However, if it's a trick question:
Either there's 4. Spades, Hearts, Diamonds, Clubs.
Or there's eight 6's, because there are two 6's on each card.
Or there's twelve 6's, because and upside-down nine is a 6.
And of fives, there are 4. La-di-da.
And of both there are 8, both 5's and 6's.
So hurrah!
Having fun playing with cards are we?
Hmm?
-Sam
2007-12-13 14:32:19
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answer #9
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answered by Sam 4
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There are 4 of everything in a standard deck of cards.
2007-12-13 14:19:23
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answer #10
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answered by Anonymous
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