1. ## Help with combinatorics

How many "words" (actually just letter combinations) can you make with the word SIMILARITIES so that there are no "words" with four consecutive I's.

If am not mistaken the total amount of "words" would be $\displaystyle 12!/(3! \cdot 2!)$

But I don't know how to calculate the amount of words that have four consecutive I's.

I would appreciate if anybody could explain this step by step.

Thank you very much!

PS: I don't really know how to use the math code, but I hope you understand what I wrote.

EDIT: I meant to write 4! as there are 4 I's

2. Originally Posted by sebasto
How many "words" (actually just letter combinations) can you make with the word SIMILARITIES so that there are no "words" with four consecutive I's.

If am not mistaken the total amount of "words" would be $\displaystyle 12!/(3! \cdot 2!)$
You are mistaken the total amount of "words" would be $\displaystyle 12!/({\color{red}4}! \cdot 2!)$
There are 4 I's not 3.

Look at the separators. $\displaystyle \left\{ {S_1 MLARTES_2 } \right\}$ the subscrips in the S's make them different.
Those create 9 places to put the I's.
So $\displaystyle {9 \choose 4}{\frac {8!} {2}}$ (divide by 2 to remove the subscripts).

3. Hello, sebasto!

How many "words" can we make with the letters in the word
SIMILARITIES, so that there are no words with four consecutive I's.

There are twelve letters, including four I's and two S's.

There are: .$\displaystyle \frac{12!}{4!\,2!} \;=\;239,500,800$ possible "words".

Duct-tape the four I's together.
Then we have nine "letters" to arrange: .$\displaystyle A, E, L, M, R, S, S, T, \boxed{IIII}$
. . They can be arranged in: .$\displaystyle \frac{9!}{2!} \;=\;181,440$ ways

Hence, there are $\displaystyle 181,440$ "words" with four consecutive I's.

Therefore, there are: .$\displaystyle 239,500,800 - 181,440 \;=\;\boxed{239,319,360}$ "words"
. . without four consecutive I's.