Combinatorial Combinations of Combinatorics

There $\displaystyle 13!$ distinct ways of choosing the letters. The letters C, O and I are repeated all twice therefore we are left with $\frac{13!}{2!\ 2!\ 2!}=13!/8=\boxed{778377600}$

The thing I did was I divided the 13! with the given options and when I divided by 778377600 I got a whole number. Thus I knew it was the right option

As C, O & I are repeated twice each, so total of 13 numbers have to be divided.

13! / 2! / 2! / 2! = 778377600

The image may confuse people. At first I thought that spaces would count given that the it shows a combination of the letters with a space in it. This would make the problem very different. You could add a space between each letter or use no spaces at all. This would also be an interesting problem but it's not what is being asked. I suggest removing the image or clarifying the question to eliminate confusion given by the image.

Ya, Easy..Every one does it like that only...wonder if we could translate this to something different...geometry maybe....?

the number of letters in the word "COMBINATORICS" is 13. then, The number of letters that is repeated is 3, and that letters are "I", "O" and "C". therefore 13 should be our numerator, BUT, put a sign "!" so that it will be on its factorial form then the denominator will be " 2!2!2! " or simply 8, because if you simplify "2!2!2!" the answer will be 8. then if you calculate it, the answer will be 778377600 :)

13!/2!2!2!

that's 13!/(2! 2! 2!)

The numer N of ways in whic can be arranged the letters of the

word:

COMBINATORICS

are the distincted permutation of 13 letters, with:

2 repetitions of letter C, 2 for O, 2 for I;

so: N = (2,2,2)P13 = 13!/(2!2!2!) = 778377600

Simple 13! for letters, but since C, O and I are there each 2 times so result is $\frac{13!}{2!.2!.2!}$ .

Very Nice Problem...

13!/2! 2! 2!