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

Question

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

A. 6
B. 30
C. 90
D. 120
E. 180

This is a permutation with indistinguishable events - repeated items. The number of different permutations of N objects, where there are N1 indistinguishable objects of style 1, N2 indistinguishable objects of style 2, ..., and Nk indistinguishable objects of style k, is = N!/(N1!*N2!* ... * Nk!). In this case, N=6; N1=2, and N2=2. This gives the formula: 6!/(2!*2!)=180

Bunuel · Accepted Answer

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.

Archit3110 · Answer

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

BrentGMATPrepNow · Answer

------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 !)]
-----------------------
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

ScottTargetTestPrep · Answer

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

SergioSantos · Answer

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

Kimberly77 · Answer

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

BrentGMATPrepNow · Answer

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.

Kimberly77 · Answer

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

BrentGMATPrepNow · Answer

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

Kimberly77 · Answer

Great thanks BrentGMATPrepNow for confirmation

Mingtrpk · Answer

If I want to solve this using combination, am I correct for using 6C2 * 4C2 * 2C2 resulting in 180 ?

Bunuel · Answer

6C2 * 4C2 * 2C2 = 90, not 180. Please check the discussion above.

Mingtrpk · Answer

Thank you so much for the reply, I check the discussion above and fully understand the idea and why I shouldn't use combination approach. Still, if I try to use 2C1 instead of 2C2, will it be a correct way of using combination in this question? or I was just super lucky that the answer match with the permutation one?

Bunuel · Answer

I don’t really follow the logic behind the formula you’re using, so I suggest reviewing the correct methods shown above.

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

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