permutation and combination

[shail4000]

Find the number of words that may be formed by permuting the letters of the word INDEPENDENCE

such that the Es do not come together.

The answer is suppose to be:

1632960

 

But, I have got something else *** shown below;

THE Es coming together can be calculated as shown:

We bind two Es, taking it as a single element; thus, we are left with 11 elements.

We now have 3Ns, 2Ds, and three Es because we have merge two Es into one.

Hence the arrangement having Es together:

11!/(3!2!3!)

The possibility space consists of:

12!/(3!2!4!)

The answer I got is 1,108,800(possibility - arrangements having Es together)

If the given answer is good, plz explain; else, figure if mine is good.

[rina777]

there are 3N, 2D and 4E.

Total number of arrangement =12!/(3!*2!*4!)

The number of arrangement where Es come together, considering 4Es as one letter=9!/(3!*2!)

 

So the number of arrangement where Es do not come together=[12!/(3!*2!*4!)] -[9!/(3!*2!)]

by calculating we get the answer=1632960

[deichgraf]

Don't you think this will be far to hard as an exercise in your real GRE?

[Riyad Mohammad]

i dont think that this is a hard type of GRE math

[tiptop]

Find the number of words that may be formed by permuting the letters of the word INDEPENDENCE

such that the Es do not come together.

 

I really hate the language they use... for my understanding they meant 2,3 or 4 Es, which makes it even more complicated.

 

Riyad, you mean it is harder than GRE or easier than hard GRE? Permutation comes up from time to time...