In how many ways could the letters in the word MINIMUM be arranged if

Question

In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?

A. 80
B. 140
C. 280
D. 300
E. 400

nick1816 · Accepted Answer

Total number of ways to arrange letters of "MINIMUM"= 7!/3!2!= 420

There are 3 ways to arrange 2 I's and 1 U: IIU/IUI/UII. But under given constraints we can choose only 1 that is IIU. Hence, divide the total numbers of ways by 3.

Total number of ways to arrange letters of "MINIMUM" if the U must not come before the 2 I's= 420/3=140

BrentGMATPrepNow · Answer

Let's first replace the I's and the U's with X's to get the word MXNXMXM

----------ASIDE--------------------------
To determine the number of ways to arrange the letters in MXNXMXM we can use 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 !)]
---------------------------

In the word MXNXMXM, 
There are  7  letters in total 
There are  3  identical M's
There are  3  identical X's
So, the total number of possible arrangements =  7 !/ 3 !)( 3 !) =  140

At this point, we must take each of our  140  arrangements of MXNXMXM and replace the X's with two I's and one U.

For example, one of the many outcomes is NM X  X MM X 
When we replace the X's with two I's and one U  AND  adhere to the rule that the U must not come before the I's, we see that there is only ONE way to do this. 
We get: NM I  I MM U

Likewise, when we take  X MM X NM X , replace the X's with two I's and one U, we get:  I MM I NM U

And when we take M X  X  X MNM, replace the X's with two I's and one U, we get: M I  I  U MNM

And so on.

So, for EACH of the  140  ways we can arrange the letters in MXNXMXM, there is ONE way to add the I's and the U.

So, the total number of ways to arrange MINIMUM =  140

Answer: B

Cheers, 
Brent

Archit3110 · Answer

MINIMUM can be arranged in ; 7!/2!*3! ; 420 ways
and IIUMNMM ; let IIU =X ; which can be arranged in 3!/2! ; 3 ways 
so total possibilities of arranging MINIMUM with the condition of U must not come before I's ; 420/3 ; 140
IMO B

EgmatQuantExpert · Answer

Solution

To find:     
 • In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?

Approach and Working Out:     
 • The total number of ways in which all the letters can be arranged =  \frac{7!}{3! * 2!}  = 420 ways
 o Since, M repeats thrice and I repeats twice
o Among these 420,  \frac{1}{3}^{rd}  will have UII order,  \frac{1}{3}^{rd}  will have IUI and  \frac{1}{3}^{rd}  will have IIU.

• Therefore, the required answer is  \frac{420}{3} = 140

Hence, the correct answer is Option B.

Answer: B

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In how many ways could the letters in the word MINIMUM be arranged if

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In how many ways could the letters in the word MINIMUM be arranged if

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