In how many ways can letters in a word CALCULUS be rearranged
0
$begingroup$
In how many ways can letters in a word CALCULUS be rearranged such that no 2 same letters stand next to each other. I’ve been thinking of Inclusion-Exclusion principle. Is there any different way to solve this task?
combinatorics inclusion-exclusion
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
$endgroup$
|
show 7 more comments
0
$begingroup$
In how many ways can letters in a word CALCULUS be rearranged such that no 2 same letters stand next to each other. I’ve been thinking of Inclusion-Exclusion principle. Is there any different way to solve this task?
combinatorics inclusion-exclusion
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
$endgroup$
1
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56
|
show 7 more comments
0
0
0
$begingroup$
In how many ways can letters in a word CALCULUS be rearranged such that no 2 same letters stand next to each other. I’ve been thinking of Inclusion-Exclusion principle. Is there any different way to solve this task?
combinatorics inclusion-exclusion
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
$endgroup$
In how many ways can letters in a word CALCULUS be rearranged such that no 2 same letters stand next to each other. I’ve been thinking of Inclusion-Exclusion principle. Is there any different way to solve this task?
combinatorics inclusion-exclusion
combinatorics inclusion-exclusion
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
edited Jan 7 at 10:30
N. F. Taussig
44.8k103358
44.8k103358
asked Jan 7 at 10:16
MichaelMichael
246
asked Jan 7 at 10:16
MichaelMichael
246
asked Jan 7 at 10:16
MichaelMichael
246
246
1
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56
|
show 7 more comments
1
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56
1
1
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56
|
show 7 more comments
0
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1
$begingroup$
Inclusion-Exclusion should work fine. You have three letters which appear twice, so you will not get that many terms in the final expansion for the rearrangements.
$endgroup$
– EuxhenH
Jan 7 at 10:28
$begingroup$
I know but I wonder if there’s any other way to do that not using sophisticated methods
$endgroup$
– Michael
Jan 7 at 11:26
$begingroup$
Not using sophisticated methods would force you to do tedious casework.
$endgroup$
– N. F. Taussig
Jan 7 at 16:02
$begingroup$
What would be the final answer then
$endgroup$
– Michael
Jan 8 at 8:52
$begingroup$
If you want additional feedback, then you should edit your post to show what you have attempted and explain where you are stuck. This tutorial explains how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Jan 8 at 10:56