How many ways can the letters in the word COMMON be arranged

THEORY FOR SUCH KIND OF PERMUTATION QUESTIONS:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is \(8!\) as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

BACK TO THE ORIGINAL QUESTION:
How many ways can the letters in the word COMMON be arranged?
A. 6
B. 30
C. 90
D. 120
E. 180

According to the above the # of permutations of 6 letters COMMON out of which 2 O's and 2 M's are identical is \(\frac{6!}{2!*2!}=180\).

Answer: E.

Hope it's clear.
_________________

total possible ways
6!/2!*2! = 180
IMO E

------ASIDE-----------------------
When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:

If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

So, for example, we can calculate the number of arrangements of the letters in MISSISSIPPI as follows:
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
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Now on to the question!

The word: COMMON:
There are 6 letters in total
There are 2 identical O's
There are 2 identical M's
So, the total number of possible arrangements = 6!/[(2!)(2!)] = 180

Answer: E

Cheers,
Brent

The total number of ways to arrange the letters in COMMON is:

6! / (2! x 2!) = (6 x 5 x 4 x 3 x 2) / (2 x 2) = 6 x 5 x 3 x 2 = 180

We use the indistinguishable permutations formula to solve this problem. If the letters were all distinct, the answer would be 6!. However, there are two O’s and two M’s, and so we divide 6! by 2! x 2! to take into account the permutations that are not distinct due to the identical O’s and M’s in the word COMMON.

Answer: E

_________________

The simplest way for this specific case:
6 letters - 6!
2 repetitions of double letters: 2!*2!
\(\frac{6!}{2!*2! }= 180\)
E

Hi BrentGMATPrepNow, to clarify question asked How many ways? So it's total combinations. Therefore not sure why we still need the denominator part of division with 2! 2!?
Thought we only need this if question asked was How many *different* ways? Did I miss something here? Thanks Brent

First, since the order in which we arrange the letters matters, this isn't a combination question.
At the same time, it's not a straightforward FCP question because we have some identical letters in the word COMMON.

I'm not sure what you mean by the *different* ways condition.
All counting questions are essentially asking you to find the total number of different outcomes.
So the word different doesn't really provide any insights into how to solve the question.

Thanks BrentGMATPrepNow. Get it. Can we presume that order matter in all words/characters count?
I mean *different* by no duplicate counts.

Yes, if we are arranging letters or digits, then the order definitely matters.

Great thanks BrentGMATPrepNow for confirmation