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combinations
- Thread starter carebear
- Start date
\(\displaystyle \binom{4}{1}\cdot\binom{12}{1}\cdot\binom{13}{1}\)
See why?.
No. It is not a fraction. C(13,1)=13. C(52,4)=270725
Quite a difference.
See why?.
Does C(52,4) reduce to C(13,1)?
No. It is not a fraction. C(13,1)=13. C(52,4)=270725
Quite a difference.
Hello, carebear!
Think about what you have written . . .
\(\displaystyle C(52,4)\) is the number of ways ANY 4 cards can be dealt from the deck.
\(\displaystyle C(48,1)\) is the number of ways 1 card can be dealt from the remaining 48 cards.
No mention is made of a "4 of a kind".
From a deck of 52 cards, how many ways can 5 cards be dealt to get "4 of a kind"?
Why isn't it C(52,4) x C(48,1)? . . . . Because it's wrong
Think about what you have written . . .
\(\displaystyle C(52,4)\) is the number of ways ANY 4 cards can be dealt from the deck.
\(\displaystyle C(48,1)\) is the number of ways 1 card can be dealt from the remaining 48 cards.
No mention is made of a "4 of a kind".
Can you please further explain C(4,1) x C(12,1) x C(13, 1) with words and not numbers? I am having trouble understanding.
Does the first part mean.....choose one of the 4 suits.....or does it relate to the 4 of a kind?
Sorry....I am soooo confused on permutations and combinations, etc..
Does the first part mean.....choose one of the 4 suits.....or does it relate to the 4 of a kind?
Sorry....I am soooo confused on permutations and combinations, etc..
C(13,1) means you are choosing 1 from the 13 values for the '4 of a kind'. You can have 4 Aces, 4 Deuces, 4 Treys, 4 Fours, etc.
C(12,1) is choosing 1 from the other 12 for the fifth card.
C(4,1) is choosing 1 from the four suits.
C(12,1) is choosing 1 from the other 12 for the fifth card.
C(4,1) is choosing 1 from the four suits.
Thank you very much.....now if the question says: How many 5 card hands contain: 3 of one kind and 2 of another...is this correct?
C(13,1) - choose a jack for example from 13 possible cards in a suit
C(4,3) - there are 4 jacks to choose from and I need to pick 3
C(12,1) - choose a seven for example, but because I have already selected a jack, i only have 12 cards remaining in that suit
C(4,2) - there are 4 sevens to choose from and I need to pick 2
I would then multiply this together.....do my explanations sound correct?
C(13,1) - choose a jack for example from 13 possible cards in a suit
C(4,3) - there are 4 jacks to choose from and I need to pick 3
C(12,1) - choose a seven for example, but because I have already selected a jack, i only have 12 cards remaining in that suit
C(4,2) - there are 4 sevens to choose from and I need to pick 2
I would then multiply this together.....do my explanations sound correct?