# How many different words can be formed using all the letters of the word GOOGOLPLEX?

How many different words can be formed using all the letters of the word GOOGOLPLEX?

I tried answering this problem and came up with the formula $n!/a!b!c!$ where $n$ in this case is 10-because it is the number of letters in the word.

$a!$ is O and in this case the letter O is repeated 3 times so $a=3$.

$b!$ is G and in this case the letter G is repeated 2 times.

and $c!$ is the other letters which are not repeated.

So the answer that i got is 302,400

Am I correct?

• L is also repeated 2 times – Shobhit Sep 3 '13 at 20:04
• okay so it is 10!/3!2!2! = 151,200? – user84466 Sep 3 '13 at 20:08
• Yes, correct you are. – Shobhit Sep 3 '13 at 20:09

Your answer is off by a factor of $2$, seems that you forgot the double letter L.

The word GOOGOLPLEX consists of the letters O (3x), G (2x), L (2x), P (1x), E (1x) and X (1x). So the number of possible words consisting of these letters is the multinomial coefficient

$$\binom{10}{3,2,2,1,1,1} = \frac{10!}{3!\cdot 2!\cdot 2!\cdot 1!\cdot 1!\cdot 1!} = 151200.$$

You are almost right.

If $a_1,a_2,\dots$ are the numbers of times different letters appear in your word, then the answer is $n!/a_1!a_2!\dots$. So, in your case the answer is $10!/3!2!2!$ (not including the $1!$'s.

The answer will be $$\frac{n!}{a!b!c!\cdots}$$ where $n$ is the number of letters in the word and $a,b,c\cdots$ denote the number of repititions of the letters in the respective word.

In this case the word is GOOGOLPLEX which has $2$ $G's$, $3$ $O's$ and $2$ $L's$ hence the answer becomes $$\frac{10!}{3!2!2!} = 151200.$$

For some reference visit : http://en.wikipedia.org/wiki/Multinomial_theorem

You have a small error. Note that the formula $\frac{n!}{a!b!\cdots}$ means that first we consider with $n!$ all permutations of the letters, repeated or not. But if we have $a$ number of same letters, then we have $a!$ ways to choose equivalent words. So $\frac{n!}{a!}$ will count number of words where we do not count words equivalent with the letter which there are $a$ same of them. If we repeat this proccess for all repeat letters we will have $\frac{n!}{a!b! \cdots}$ is the answer.