7 reindeer in a single file line, dasher cannot be next to prancer.
$begingroup$
overall, I can see that there would be 7! arrangements, but I don't know how to remove the instances where Dasher is next to Prancer. I can think of 12 arrangements of which dasher is next to prancer.
combinatorics permutations
asked Dec 5 '18 at 15:58
nessness
375
$endgroup$
add a comment |
$begingroup$
overall, I can see that there would be 7! arrangements, but I don't know how to remove the instances where Dasher is next to Prancer. I can think of 12 arrangements of which dasher is next to prancer.
combinatorics permutations
asked Dec 5 '18 at 15:58
nessness
375
$endgroup$
add a comment |
$begingroup$
overall, I can see that there would be 7! arrangements, but I don't know how to remove the instances where Dasher is next to Prancer. I can think of 12 arrangements of which dasher is next to prancer.
combinatorics permutations
asked Dec 5 '18 at 15:58
nessness
375
$endgroup$
overall, I can see that there would be 7! arrangements, but I don't know how to remove the instances where Dasher is next to Prancer. I can think of 12 arrangements of which dasher is next to prancer.
combinatorics permutations
combinatorics permutations
asked Dec 5 '18 at 15:58
nessness
375
asked Dec 5 '18 at 15:58
nessness
375
asked Dec 5 '18 at 15:58
nessness
375
asked Dec 5 '18 at 15:58
nessness
375
asked Dec 5 '18 at 15:58
nessness
375
375
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
HINT: To obtain the number of arrangements where Dasher and Prancer are together, view them as a single unit. Think of all the ways that Dancer, Vixen, Comet, Cupid, Blitzen, and "DasherPrancer" can be arranged. Note that an arrangement where Dasher is on Prancer's left is distinct from an arrangement where Dasher is on Prancer's right, so you'll have to eliminate both sets of arrangements from your total count.
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
$endgroup$
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
add a comment |
$begingroup$
There are two possibilities of Dasher being on the end. Each of those has 5 possibilities for Prancer and then 5! options for the other 5 reindeer, for a total of 1200 possibilities with Dasher on the end. There are five possibilities of Dasher not on an end. Each has four possibilities for Prancer and then 5! options for the other reindeer, for a total of 2400 with Dasher somewhere in the middle. 3600 arrangements total
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT: To obtain the number of arrangements where Dasher and Prancer are together, view them as a single unit. Think of all the ways that Dancer, Vixen, Comet, Cupid, Blitzen, and "DasherPrancer" can be arranged. Note that an arrangement where Dasher is on Prancer's left is distinct from an arrangement where Dasher is on Prancer's right, so you'll have to eliminate both sets of arrangements from your total count.
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
$endgroup$
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
add a comment |
$begingroup$
HINT: To obtain the number of arrangements where Dasher and Prancer are together, view them as a single unit. Think of all the ways that Dancer, Vixen, Comet, Cupid, Blitzen, and "DasherPrancer" can be arranged. Note that an arrangement where Dasher is on Prancer's left is distinct from an arrangement where Dasher is on Prancer's right, so you'll have to eliminate both sets of arrangements from your total count.
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
$endgroup$
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
add a comment |
$begingroup$
HINT: To obtain the number of arrangements where Dasher and Prancer are together, view them as a single unit. Think of all the ways that Dancer, Vixen, Comet, Cupid, Blitzen, and "DasherPrancer" can be arranged. Note that an arrangement where Dasher is on Prancer's left is distinct from an arrangement where Dasher is on Prancer's right, so you'll have to eliminate both sets of arrangements from your total count.
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
$endgroup$
HINT: To obtain the number of arrangements where Dasher and Prancer are together, view them as a single unit. Think of all the ways that Dancer, Vixen, Comet, Cupid, Blitzen, and "DasherPrancer" can be arranged. Note that an arrangement where Dasher is on Prancer's left is distinct from an arrangement where Dasher is on Prancer's right, so you'll have to eliminate both sets of arrangements from your total count.
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
answered Dec 5 '18 at 16:05
Michael SeifertMichael Seifert
4,757624
4,757624
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
add a comment |
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
$begingroup$
I want to use 2*6!, but 7!/(2*6!) seems off.
$endgroup$
– ness
Dec 5 '18 at 16:09
1
1
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
@ness: You're removing one set of permutations from another, not taking some kind of equivalence class. Think of it this way: if you have six pets at home, and two of them are birds, how many are not birds? The answer isn't 6/2.
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:12
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
$begingroup$
7!-(2*6!)=3600 gives a much more reasonable answer, but I can't quite see why we use subtraction instead of division.
$endgroup$
– ness
Dec 5 '18 at 16:20
1
1
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
@ness: Imagine making a list of the possible arrangements of the reindeer, assuming you don't care about where Dasher and Prancer are. Now make another list of all the arrangements where Dasher is next to Prancer. Now take any arrangement in the second list and cross it off of the first list, so that the first list now only has arrangements where Dancer & Prancer aren't next to each other. How many arrangements remain on the first list?
$endgroup$
– Michael Seifert
Dec 5 '18 at 16:24
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
$begingroup$
thanks so much! :)
$endgroup$
– ness
Dec 5 '18 at 17:13
add a comment |
$begingroup$
There are two possibilities of Dasher being on the end. Each of those has 5 possibilities for Prancer and then 5! options for the other 5 reindeer, for a total of 1200 possibilities with Dasher on the end. There are five possibilities of Dasher not on an end. Each has four possibilities for Prancer and then 5! options for the other reindeer, for a total of 2400 with Dasher somewhere in the middle. 3600 arrangements total
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
There are two possibilities of Dasher being on the end. Each of those has 5 possibilities for Prancer and then 5! options for the other 5 reindeer, for a total of 1200 possibilities with Dasher on the end. There are five possibilities of Dasher not on an end. Each has four possibilities for Prancer and then 5! options for the other reindeer, for a total of 2400 with Dasher somewhere in the middle. 3600 arrangements total
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
$endgroup$
add a comment |
$begingroup$
There are two possibilities of Dasher being on the end. Each of those has 5 possibilities for Prancer and then 5! options for the other 5 reindeer, for a total of 1200 possibilities with Dasher on the end. There are five possibilities of Dasher not on an end. Each has four possibilities for Prancer and then 5! options for the other reindeer, for a total of 2400 with Dasher somewhere in the middle. 3600 arrangements total
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
$endgroup$
There are two possibilities of Dasher being on the end. Each of those has 5 possibilities for Prancer and then 5! options for the other 5 reindeer, for a total of 1200 possibilities with Dasher on the end. There are five possibilities of Dasher not on an end. Each has four possibilities for Prancer and then 5! options for the other reindeer, for a total of 2400 with Dasher somewhere in the middle. 3600 arrangements total
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
answered Dec 5 '18 at 16:02
MoKo19MoKo19
1914
1914
add a comment |
add a comment |
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