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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

5

3

$begingroup$

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

share|cite|improve this question

edited 8 hours ago

King_Iverson

asked 8 hours ago

King_IversonKing_Iverson

5633 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
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are correct. it's 4!4!2!. Editing
  $endgroup$
  – King_Iverson
  8 hours ago
  

  
  
  
  

add a comment
 | 

5

3

$begingroup$

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

share|cite|improve this question

edited 8 hours ago

King_Iverson

asked 8 hours ago

King_IversonKing_Iverson

5633 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
  8 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

share|cite|improve this question

edited 8 hours ago

King_Iverson

asked 8 hours ago

King_IversonKing_Iverson

5633 bronze badges

$endgroup$

share|cite|improve this question

edited 8 hours ago

King_Iverson

asked 8 hours ago

King_IversonKing_Iverson

5633 bronze badges

share|cite|improve this question

edited 8 hours ago

King_Iverson

asked 8 hours ago

King_IversonKing_Iverson

5633 bronze badges

asked 8 hours ago

King_IversonKing_Iverson

5633 bronze badges

asked 8 hours ago

King_IversonKing_Iverson

5633 bronze badges

1 Answer
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oldest

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7

$begingroup$

There is a way to do this, actually.

First, order the letters MSSSSPP. This can be done in $displaystyle frac7!4!2!=105$ 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 $displaystyle frac11!4!4!2!-7350=34650-7350=27300$ ways.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! Great explanation!
  $endgroup$
  – King_Iverson
  8 hours ago
  

  
  
  
  

add a comment
 | 

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7

$begingroup$

There is a way to do this, actually.

First, order the letters MSSSSPP. This can be done in $displaystyle frac7!4!2!=105$ 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 $displaystyle frac11!4!4!2!-7350=34650-7350=27300$ ways.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks! Great explanation!
  $endgroup$
  – King_Iverson
  8 hours ago
  

  
  
  
  

add a comment
 | 

7

$begingroup$

There is a way to do this, actually.

First, order the letters MSSSSPP. This can be done in $displaystyle frac7!4!2!=105$ 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 $displaystyle frac11!4!4!2!-7350=34650-7350=27300$ ways.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 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 $displaystyle frac7!4!2!=105$ 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 $displaystyle frac11!4!4!2!-7350=34650-7350=27300$ ways.

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges

answered 8 hours ago

Saketh MalyalaSaketh Malyala

11.7k1717 silver badges3939 bronze badges