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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
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.
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.
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.
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
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?
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.
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
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.
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 _________________
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
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?