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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
$begingroup$
How many anagrams of the word mississippi are there that have at least two consecutive i’s?
My approach was: Finding the total amount of anagrams (11! / 4! 4! 2!).
And then to calculate the amount of options without 2 consecutive i's.
In the end, to find the difference between them.
However, the result felt me to big intuitively.
Do I miss something?
Thanks
combinatorics
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
$endgroup$
add a comment
|
$begingroup$
How many anagrams of the word mississippi are there that have at least two consecutive i’s?
My approach was: Finding the total amount of anagrams (11! / 4! 4! 2!).
And then to calculate the amount of options without 2 consecutive i's.
In the end, to find the difference between them.
However, the result felt me to big intuitively.
Do I miss something?
Thanks
combinatorics
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
$endgroup$
$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
$begingroup$
How many anagrams of the word mississippi are there that have at least two consecutive i’s?
My approach was: Finding the total amount of anagrams (11! / 4! 4! 2!).
And then to calculate the amount of options without 2 consecutive i's.
In the end, to find the difference between them.
However, the result felt me to big intuitively.
Do I miss something?
Thanks
combinatorics
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
$endgroup$
How many anagrams of the word mississippi are there that have at least two consecutive i’s?
My approach was: Finding the total amount of anagrams (11! / 4! 4! 2!).
And then to calculate the amount of options without 2 consecutive i's.
In the end, to find the difference between them.
However, the result felt me to big intuitively.
Do I miss something?
Thanks
combinatorics
combinatorics
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
edited 8 hours ago
King_Iverson
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
asked 9 hours ago
King_IversonKing_Iverson
563 bronze badges
563 bronze badges
$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago
$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
1 Answer
1
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$begingroup$
There is a way to do this, actually.
First, order the letters MSSSSPP. This can be done in $displaystyle frac{7!}{4!2!}=105$ ways.
Now, you have $8$ spaces between those letters to possibly insert I's (ends included).
Now, choose $4$ of those $8$ spaces to insert an $I$.
This gives a total of $105cdot 70=7350$ ways.
Now, we are looking for the complement, so our answer is $displaystyle frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
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1 Answer
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1 Answer
1
active
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$begingroup$
There is a way to do this, actually.
First, order the letters MSSSSPP. This can be done in $displaystyle frac{7!}{4!2!}=105$ ways.
Now, you have $8$ spaces between those letters to possibly insert I's (ends included).
Now, choose $4$ of those $8$ spaces to insert an $I$.
This gives a total of $105cdot 70=7350$ ways.
Now, we are looking for the complement, so our answer is $displaystyle frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
$begingroup$
There is a way to do this, actually.
First, order the letters MSSSSPP. This can be done in $displaystyle frac{7!}{4!2!}=105$ ways.
Now, you have $8$ spaces between those letters to possibly insert I's (ends included).
Now, choose $4$ of those $8$ spaces to insert an $I$.
This gives a total of $105cdot 70=7350$ ways.
Now, we are looking for the complement, so our answer is $displaystyle frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
$endgroup$
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
$begingroup$
There is a way to do this, actually.
First, order the letters MSSSSPP. This can be done in $displaystyle frac{7!}{4!2!}=105$ ways.
Now, you have $8$ spaces between those letters to possibly insert I's (ends included).
Now, choose $4$ of those $8$ spaces to insert an $I$.
This gives a total of $105cdot 70=7350$ ways.
Now, we are looking for the complement, so our answer is $displaystyle frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
$endgroup$
There is a way to do this, actually.
First, order the letters MSSSSPP. This can be done in $displaystyle frac{7!}{4!2!}=105$ ways.
Now, you have $8$ spaces between those letters to possibly insert I's (ends included).
Now, choose $4$ of those $8$ spaces to insert an $I$.
This gives a total of $105cdot 70=7350$ ways.
Now, we are looking for the complement, so our answer is $displaystyle frac{11!}{4!4!2!}-7350=34650-7350=27300$ ways.
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
answered 9 hours ago
Saketh MalyalaSaketh Malyala
11.7k17 silver badges39 bronze badges
11.7k17 silver badges39 bronze badges
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
1
1
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
$begingroup$
Thanks! Great explanation!
$endgroup$
– King_Iverson
8 hours ago
add a comment
|
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$begingroup$
Well, you have a typo. You mean ${11!over 4!4!2!}$ Is this what's making your answer too big?
$endgroup$
– saulspatz
9 hours ago
$begingroup$
You are correct. it's 4!4!2!. Editing
$endgroup$
– King_Iverson
8 hours ago