Reason for difference in number of five-card hands.
Find the number of five-card hands dealt from a deck of $52$ cards, s.t. there is one pair (two cards of one denomination), a third card of a different denomination, a fourth card of a third different denomination, and a fifth card of a fourth different denomination.
My approach is:
There is application of product principle for all the sub-cases (& in the sub-cases as well).
(i) There are $binom{13}{1}$ ways to get one denomination, then choose two cards out of four suits by $binom{4}{2}$.
(ii) Further, the third card can be chosen in $binom{12}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iii) The fourth card can be chosen in $binom{11}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iv) Further, the fifth card can be chosen in $binom{10}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
The answer is : $(binom{13}{1}*binom{4}{2})*(binom{12}{1}*4)*(binom{11}{1}*4)*(binom{10}{1}*4)$
But the answer is given by :
$binom{13}{1}* binom{4}{2}* binom{12}{3}* binom{4}{1}^3$
The both approaches given differ by a factor of $3$.
My approach yields higher value by effectively yielding a permutation of $3$ cards from $12$; while the answer yields a combination of them. But, am unclear how my approach is wrong in taking individual choices of :$binom{12}{1}, binom{11}{1}, binom{10}{1}$.
combinatorics
asked Nov 19 '18 at 5:13
jiten
1,2411413
add a comment |
Find the number of five-card hands dealt from a deck of $52$ cards, s.t. there is one pair (two cards of one denomination), a third card of a different denomination, a fourth card of a third different denomination, and a fifth card of a fourth different denomination.
My approach is:
There is application of product principle for all the sub-cases (& in the sub-cases as well).
(i) There are $binom{13}{1}$ ways to get one denomination, then choose two cards out of four suits by $binom{4}{2}$.
(ii) Further, the third card can be chosen in $binom{12}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iii) The fourth card can be chosen in $binom{11}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iv) Further, the fifth card can be chosen in $binom{10}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
The answer is : $(binom{13}{1}*binom{4}{2})*(binom{12}{1}*4)*(binom{11}{1}*4)*(binom{10}{1}*4)$
But the answer is given by :
$binom{13}{1}* binom{4}{2}* binom{12}{3}* binom{4}{1}^3$
The both approaches given differ by a factor of $3$.
My approach yields higher value by effectively yielding a permutation of $3$ cards from $12$; while the answer yields a combination of them. But, am unclear how my approach is wrong in taking individual choices of :$binom{12}{1}, binom{11}{1}, binom{10}{1}$.
combinatorics
asked Nov 19 '18 at 5:13
jiten
1,2411413
add a comment |
Find the number of five-card hands dealt from a deck of $52$ cards, s.t. there is one pair (two cards of one denomination), a third card of a different denomination, a fourth card of a third different denomination, and a fifth card of a fourth different denomination.
My approach is:
There is application of product principle for all the sub-cases (& in the sub-cases as well).
(i) There are $binom{13}{1}$ ways to get one denomination, then choose two cards out of four suits by $binom{4}{2}$.
(ii) Further, the third card can be chosen in $binom{12}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iii) The fourth card can be chosen in $binom{11}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iv) Further, the fifth card can be chosen in $binom{10}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
The answer is : $(binom{13}{1}*binom{4}{2})*(binom{12}{1}*4)*(binom{11}{1}*4)*(binom{10}{1}*4)$
But the answer is given by :
$binom{13}{1}* binom{4}{2}* binom{12}{3}* binom{4}{1}^3$
The both approaches given differ by a factor of $3$.
My approach yields higher value by effectively yielding a permutation of $3$ cards from $12$; while the answer yields a combination of them. But, am unclear how my approach is wrong in taking individual choices of :$binom{12}{1}, binom{11}{1}, binom{10}{1}$.
combinatorics
asked Nov 19 '18 at 5:13
jiten
1,2411413
Find the number of five-card hands dealt from a deck of $52$ cards, s.t. there is one pair (two cards of one denomination), a third card of a different denomination, a fourth card of a third different denomination, and a fifth card of a fourth different denomination.
My approach is:
There is application of product principle for all the sub-cases (& in the sub-cases as well).
(i) There are $binom{13}{1}$ ways to get one denomination, then choose two cards out of four suits by $binom{4}{2}$.
(ii) Further, the third card can be chosen in $binom{12}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iii) The fourth card can be chosen in $binom{11}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
(iv) Further, the fifth card can be chosen in $binom{10}{1}$ ways, with a particular card chosen in $binom{4}{1}$ ways.
The answer is : $(binom{13}{1}*binom{4}{2})*(binom{12}{1}*4)*(binom{11}{1}*4)*(binom{10}{1}*4)$
But the answer is given by :
$binom{13}{1}* binom{4}{2}* binom{12}{3}* binom{4}{1}^3$
The both approaches given differ by a factor of $3$.
My approach yields higher value by effectively yielding a permutation of $3$ cards from $12$; while the answer yields a combination of them. But, am unclear how my approach is wrong in taking individual choices of :$binom{12}{1}, binom{11}{1}, binom{10}{1}$.
combinatorics
combinatorics
asked Nov 19 '18 at 5:13
jiten
1,2411413
asked Nov 19 '18 at 5:13
jiten
1,2411413
edited Nov 19 '18 at 5:19
asked Nov 19 '18 at 5:13
jiten
1,2411413
asked Nov 19 '18 at 5:13
jiten
1,2411413
asked Nov 19 '18 at 5:13
jiten
1,2411413
1,2411413
add a comment |
add a comment |
1 Answer
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They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
add a comment |
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They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
add a comment |
They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
add a comment |
They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
They differ by a factor of $3!=6$, not $3$. You choose the pair first, then choose the other three cards in order. It doesn't matter what order the other cards come in, which is the factor $3!$. You count AS AH 2D 3D 4D as different from AS AH 3D 4D 2D but should not.
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
answered Nov 19 '18 at 5:26
Ross Millikan
292k23196371
292k23196371
add a comment |
add a comment |
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