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PS : Counting Methods..MINIMUM [#permalink]
 01 Oct 2003, 16:06
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?
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M 3
I 2
N 1
U 1
total number of arranging the letters is 7!/(3!*2!*1!*1!)=420
consider U and Is among other letters.
Possible variations: I I U, I U I, U I I. I think the three combinations are equally probable, and we need the first one. 420/3=140
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Praet,
whats the official answer?
i'm getting 400.....
reasoning:-- total ways of arranging is 420.
lets keep the 2 I's at the end.
so total ways of arranging the remaining 5,among which there is an I =
5!/3!1!1! i.e 20 ways.
therefore the number of ways to arrange where the U doesnt come after the I's = 420 - 20 ..ways.
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I am getting 80.
There are 7 location
1 2 3 4 5 6 7
U can not be in the first 2 positions. Because in that case I will definately follow U and violate the condition.
We need to calculate the number of ways to arrange the letter when U is in 3rd, 4th, ...7th position.
Consider U is in the 3rd place. That means we can select only M and N for positions 4 to 7 and 2 I's in the first two positions. Number of ways for this = 4! / 3! = 4
Similarly if U is in the 4th position, the only letters to the left could be (I's and M) OR (I's and N).
If the left side letters are I's and M, Number of ways in this case = (Number of ways for the positions to the left of U) * (Number of ways for the positions to the right of U)
= (3!/2! X 3!/2!)
Sililarly, if the letters in the left are I's and N = 3!/2! X 1
fOR U to be in the 4th position , total ways = (3!/2! X 3!/2!) + 3!/2! X 1
Similarly find out the number of ways for all the positions of U and sum them up. That will give 80.
Do not know if I am correct.
praetorian123, please let us know the answer.
Thanks
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I go with 140 too. By placing U at all the places following the third one, placing the two I's before them and then finding all possible permutations for the remaining places.
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stolyar wrote: M 3
I 2
N 1
U 1
total number of arranging the letters is 7!/(3!*2!*1!*1!)=420
consider U and Is among other letters.
Possible variations: I I U, I U I, U I I. I think the three combinations are equally probable, and we need the first one. 420/3=140
I concur that this is the simplest way to solve this.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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M 3
I 2
N 1
U 1
total number of arranging the letters is 7!/(3!*2!*1!*1!)=420
consider U and Is among other letters.
Possible variations: I I U, I U I, U I I. I think the three combinations are equally probable, and we need the first one. 420/3=140
---
Would you explain the 3!*2!*1!*1! part of this equation?
I understand where you got the numbers, but why do we divide by this value?
Thanks,
CJ
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CEO
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stolyar wrote: M 3
I 2
N 1
U 1
total number of arranging the letters is 7!/(3!*2!*1!*1!)=420
consider U and Is among other letters.
Possible variations: I I U, I U I, U I I. I think the three combinations are equally probable, and we need the first one. 420/3=140
140 is correct, stolyar explain why you divide by 3.
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GMAT Instructor
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praetorian123 wrote: stolyar wrote: M 3
I 2
N 1
U 1
total number of arranging the letters is 7!/(3!*2!*1!*1!)=420
consider U and Is among other letters.
Possible variations: I I U, I U I, U I I. I think the three combinations are equally probable, and we need the first one. 420/3=140 140 is correct, stolyar explain why you divide by 3. Simple logic. In all of the arrangements, either the U is after the two Is, before the 2 Is, or between the 2 Is. All of them are equally likely so the one we want happens 1/3 of the time.
_________________
Best,
AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993
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