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# In how many ways could the letters in the word MINIMUM be ar

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CEO
Joined: 15 Aug 2003
Posts: 3454
In how many ways could the letters in the word MINIMUM be ar [#permalink]

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01 Oct 2003, 16:06
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In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?
SVP
Joined: 03 Feb 2003
Posts: 1604

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03 Oct 2003, 05:08
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
Intern
Joined: 13 Sep 2003
Posts: 43
Location: US

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01 Nov 2003, 15:21
Praet,
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.
Manager
Joined: 11 Mar 2003
Posts: 54
Location: Chicago

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01 Nov 2003, 16:13
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.

Thanks
Intern
Joined: 10 Oct 2003
Posts: 45
Location: Finland

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02 Nov 2003, 01:41
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.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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29 Nov 2003, 01:14
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

Manager
Joined: 22 Nov 2003
Posts: 54
Location: New Orleans

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04 Dec 2003, 10:23
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
CEO
Joined: 15 Aug 2003
Posts: 3454

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04 Dec 2003, 15:29
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.
GMAT Instructor
Joined: 07 Jul 2003
Posts: 770
Location: New York NY 10024
Schools: Haas, MFE; Anderson, MBA; USC, MSEE

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04 Dec 2003, 21:47
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

Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There
In how many ways could the letters in the word MINIMUM [#permalink]

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15 May 2004, 16:47
In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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15 May 2004, 17:00
hallelujah1234 wrote:
In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?

7!-(combinations where U come before I)
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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15 May 2004, 17:07
mirhaque wrote:
hallelujah1234 wrote:
In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?

7!-(combinations where U come before I)

And what is within the bracket is the tougher part to calculate. Do you want to try?
_________________

Best Regards,

Paul

Senior Manager
Joined: 02 Feb 2004
Posts: 344

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15 May 2004, 17:18
Paul wrote:
mirhaque wrote:
hallelujah1234 wrote:
In how many ways could the letters in the word MINIMUM be arranged if the U must not come before the I's?

7!-(combinations where U come before I)

And what is within the bracket is the tougher part to calculate. Do you want to try?

ahhhhhhhhhhhh! nah!

it's been only a week I learned combination. Can't run before I learn to walk.
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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15 May 2004, 17:20
hallelujah1234 wrote:
Total # words = 7!/(3!2!), not 7!

can you explain this as humanely as possible
Senior Manager
Joined: 02 Mar 2004
Posts: 327
Location: There

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15 May 2004, 17:25
mirhaque wrote:
hallelujah1234 wrote:
Total # words = 7!/(3!2!), not 7! :-)

can you explain this as humanely as possible :panel

Replace 3.M and 2.N with M1, M2, and M2 and N1, and N2 respectively.

So, we can have 7! different words. However, M1 = M2 = M3, and N1 = N2. Hence, we need to divide the total words by 3! and 2! respectively in order to knock dummies off of the list, because
{M1M2M3, M1M3M2, M2M3M1, M2M1M3, M3M1M2, M3M2M1} --> MMM
3! to 1 map.

Similarly {N1N2, N2N1} --> NN (2! to 1 map)
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4288

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15 May 2004, 18:09
Total number of ways to arrange letters of MINIMUM to form distinct words: 7!/2!*3! = 420

Unfavorable outcomes are when U is in front of I:
(UI)-X-X-X-X-X --> 6!
Similar outcomes with M's interchanged: 3!
Total unfavorable outcomes: 6!/3! = 120

Total # of ways to arrange letters of MINIMUM such that U does not come before I: 420-120 = 300
_________________

Best Regards,

Paul

Senior Manager
Joined: 02 Feb 2004
Posts: 344

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15 May 2004, 18:23
hallelujah1234 wrote:
mirhaque wrote:
hallelujah1234 wrote:
Total # words = 7!/(3!2!), not 7!

can you explain this as humanely as possible

Replace 3.M and 2.N with M1, M2, and M2 and N1, and N2 respectively.

So, we can have 7! different words. However, M1 = M2 = M3, and N1 = N2. Hence, we need to divide the total words by 3! and 2! respectively in order to knock dummies off of the list, because
{M1M2M3, M1M3M2, M2M3M1, M2M1M3, M3M1M2, M3M2M1} --> MMM
3! to 1 map.

Similarly {N1N2, N2N1} --> NN (2! to 1 map)

why not knock off dummies for "I"s as well. there are two I's
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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17 May 2004, 05:45
Paul wrote:
Total number of ways to arrange letters of MINIMUM to form distinct words: 7!/2!*3! = 420

Unfavorable outcomes are when U is in front of I:
(UI)-X-X-X-X-X --> 6!
Similar outcomes with M's interchanged: 3!
Total unfavorable outcomes: 6!/3! = 120

Total # of ways to arrange letters of MINIMUM such that U does not come before I: 420-120 = 300

Problem with this
Quote:
(UI)-X-X-X-X-X --> 6!
is: there are two "I"s & with this combination the second "I" will come before "U". However, if you assume "UII" as one, that eliminates that problem but does not count all combinations where other letters could be between the two "I"s but not before "U". What is the solution Halle?
Senior Manager
Joined: 23 Sep 2003
Posts: 293
Location: US

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17 May 2004, 06:10
To figure out ways U can come before the I's, think of the I's as a unit.

U _ _ _ _ I -------> there are 4! for the positions in the middle.

There are also 5 ways U can come before I. The above and:

_ U _ _ _ I
_ _ U _ _ I
_ _ _ U _ I
_ _ _ _ U I

= 5(4!) = 120

(7!/3!2!) - 5(4!) = 420 - 120 = 300
Senior Manager
Joined: 02 Feb 2004
Posts: 344

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17 May 2004, 08:16
ndidi204 wrote:
To figure out ways U can come before the I's, think of the I's as a unit.

U _ _ _ _ I -------> there are 4! for the positions in the middle.

There are also 5 ways U can come before I. The above and:

_ U _ _ _ I
_ _ U _ _ I
_ _ _ U _ I
_ _ _ _ U I

= 5(4!) = 120

(7!/3!2!) - 5(4!) = 420 - 120 = 300

but you are not counting other letters that can come betwee the two I's
17 May 2004, 08:16

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