Counting pairs of animalsUnordered outcomes (counting)counting on $4$ pairs of glovesMaximum unique pairings between two lists of repeated itemsNumber of Possible Pairs from Two Setshow many permutation without fixed points from 2n to NxNFeller's Probability - Counting problem of pairs of shoes.Help distinguishing between factorials, $^nC_r$ and $^nP_r$Combinatorics card counting question.Number of possible permutations of three pairs of socks, one blue, one black, and one white?Counting with Permutations
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Counting pairs of animals
Unordered outcomes (counting)counting on $4$ pairs of glovesMaximum unique pairings between two lists of repeated itemsNumber of Possible Pairs from Two Setshow many permutation without fixed points from 2n to NxNFeller's Probability - Counting problem of pairs of shoes.Help distinguishing between factorials, $^nC_r$ and $^nP_r$Combinatorics card counting question.Number of possible permutations of three pairs of socks, one blue, one black, and one white?Counting with Permutations
$begingroup$
Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)
The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?
combinatorics
edited Mar 22 at 2:10
Eevee Trainer
10k31740
asked Mar 22 at 1:59
sumisumi
654
$endgroup$
add a comment |
$begingroup$
Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)
The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?
combinatorics
edited Mar 22 at 2:10
Eevee Trainer
10k31740
asked Mar 22 at 1:59
sumisumi
654
$endgroup$
add a comment |
$begingroup$
Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)
The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?
combinatorics
edited Mar 22 at 2:10
Eevee Trainer
10k31740
asked Mar 22 at 1:59
sumisumi
654
$endgroup$
Old MacDonald has $5$ chickens, $4$ donkeys, and $7$ emus. How many ways can he pair up the animals so that every pair consists of animals of different species? (The order of the animals within each pair does not matter, and the order among the pairs does not matter. Assume that all animals are distinguishable.)
The answer I have currently is $83$ since there are $5*4=20$ possible pairs for (Chicken, Donkey), $5*7=35$ possible pairs for (Chicken, Emu), $4*7=28$ possible pairs for (Donkey, Emu). Then I added them up: $20+35+28=83$. Is this correct?
combinatorics
combinatorics
edited Mar 22 at 2:10
Eevee Trainer
10k31740
asked Mar 22 at 1:59
sumisumi
654
edited Mar 22 at 2:10
Eevee Trainer
10k31740
asked Mar 22 at 1:59
sumisumi
654
edited Mar 22 at 2:10
Eevee Trainer
10k31740
edited Mar 22 at 2:10
Eevee Trainer
10k31740
edited Mar 22 at 2:10
Eevee Trainer
10k31740
10k31740
asked Mar 22 at 1:59
sumisumi
654
asked Mar 22 at 1:59
sumisumi
654
asked Mar 22 at 1:59
sumisumi
654
654
add a comment |
add a comment |
2 Answers
2
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$begingroup$
Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..
answered Mar 22 at 2:17
WW1WW1
7,3501712
$endgroup$
add a comment |
$begingroup$
Yup, your solution is correct.
Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
$endgroup$
add a comment |
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2 Answers
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2 Answers
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active
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active
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$begingroup$
Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..
answered Mar 22 at 2:17
WW1WW1
7,3501712
$endgroup$
add a comment |
$begingroup$
Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..
answered Mar 22 at 2:17
WW1WW1
7,3501712
$endgroup$
add a comment |
$begingroup$
Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..
answered Mar 22 at 2:17
WW1WW1
7,3501712
$endgroup$
Correct - you get the same thing by counting all possible pairs and subtracting the ones with two of the same type ..
answered Mar 22 at 2:17
WW1WW1
7,3501712
answered Mar 22 at 2:17
WW1WW1
7,3501712
answered Mar 22 at 2:17
WW1WW1
7,3501712
answered Mar 22 at 2:17
WW1WW1
7,3501712
7,3501712
add a comment |
add a comment |
$begingroup$
Yup, your solution is correct.
Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
$endgroup$
add a comment |
$begingroup$
Yup, your solution is correct.
Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
$endgroup$
add a comment |
$begingroup$
Yup, your solution is correct.
Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
$endgroup$
Yup, your solution is correct.
Making this a community wiki post since it's not like I have anything to add. If anyone else does, they're welcome to, though!
answered Mar 22 at 2:08
community wiki
Eevee Trainer
add a comment |
add a comment |
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