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I'm going to try to collect some P and C type questions and

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 [#permalink] New post 13 Dec 2006, 15:07
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Rayn wrote:
How many different 6-letter arrangements ca nbe formed from the word THIRST if T must always immediately precede H?


You have T H I R S T (am not suggesting you go for a beer, it´s prep time guys :lol: )

Now, T must come right before H, thus you have any of these combos:

T H X X X X

X T H X X X

X X T H X X

X X X T H X

X X X X T H

As it´s not stated, you don´t have to pay attention whether "1st T" or "2nd T" come before or after the other, just concentrate on having a T H paired and 4 other letters to arrange. Therefore, the answer would be

5* 4! = 5! = 120.
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 [#permalink] New post 26 Jan 2007, 11:19
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Quote:
Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21


I have a question about my approach. Your approach is better and fast here, however I was trying to solve this with my approach as follows and getting a different answer, I am wondering what am I missing? Can u please correct me.

P(Choosing 2 Vowels) = 5/9*4/8
P(Choosing 2 Consonants) = 4/7*3/6 (assuming no replacement)

Combine probability is = 5/9*4/8*4/7*3/6 = 5/9*1/2*4/7*1/2

Now number of ways to arrange above is 4! and hence final answer should be 4!*5/9*1/2*4/7*1/2 = 40/21

I understand that this approach is cumbersome but what is that I am doing wrong here? Why am I not getting 10/21?
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 [#permalink] New post 27 Jan 2007, 18:10
Probability should be between 0 and 1 isn't it? 40/21 wouldn't be right.

I think your problem is you shouldn't multiply 4!, since you are using permutation in your first step. In other words, they are already ordered. You just need to pick what position you want to insert the vowls relative to the consonants. There are 6 possibilities. (VVCC, VCVC, VCCV, CVVC, CVCV,CCVV).
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Re: Letters and Words [#permalink] New post 22 Feb 2007, 05:26
HongHu wrote:
Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21



Can anyone please explain why we can't count the probability like this?:

5/9 x 4/8 x 4/7 x 3/6 (VxVxCxC) = 5/63
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 [#permalink] New post 22 Feb 2007, 05:36
HongHu wrote:
There are 6 possibilities. (VVCC, VCVC, VCCV, CVVC, CVCV,CCVV).


All possibilities separately give us the probability 240/3024 = 5/63. Why should we multiply this by 6?
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Re: Letters and Words [#permalink] New post 24 Feb 2007, 07:15
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AdrianG wrote:
HongHu wrote:
Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21



Can anyone please explain why we can't count the probability like this?:

5/9 x 4/8 x 4/7 x 3/6 (VxVxCxC) = 5/63


The way you are doing it, you've given it a specific order.

For example, if there are 2 red balls and 2 white balls, what is the probability of getting 1 white ball and 1 red ball out in the first two draws? It's C(2,1)*C(2,1)/C(4,2) = 2/3.
You could write it out to verify:
R1R2
R1W1
R1W2
R2R1
R2W1
R2W2
W1R1
W1R2
W1W2
W2R1
W2R2
W2W1

But using your method you will have 2/4*2/3=1/3. Your problem is you have specified that you will draw a white ball first, and then red ball second. However there is also the possibility of drawing a red ball first, and then a white ball second. So you need to add another 1/3 to your result to get 2/3.
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 [#permalink] New post 01 Mar 2007, 16:33
what is the meaning of ^ sign here?
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 [#permalink] New post 01 Mar 2007, 23:33
amd08 wrote:
what is the meaning of ^ sign here?


First, welcome to the GMATClub :)

It does mean power : "3^2" = "3 power 2" = 3*3 = 9
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Tow Combination Problems [#permalink] New post 15 Mar 2007, 09:20
Guys, please help me with the following problems:

1. A committee of 3 people is to be chosen from four 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A.28 B.32 C.34 D.41 E.56


2. A certain office supply store stocks 2 sizes of self-stick notepads, each in 4 colors: blue, green yellow, or pink. The store packs the notepads in packages that contain EITHER 3 notepads of the same size and the same color OR 3 notepads of the same size and the different colors. If the order in which the colors are packed is not considered, how many different packages of the types described above are possible?

A.6 B.8 C.16 D.24 E.32
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 [#permalink] New post 20 Sep 2007, 19:27
HongHu wrote:
gmat2me2 wrote:
Question: in the above example, why isn't total outcome = 4^3?

I'm not sure if you are talking about the AUSTRALIA problem. In that problem we need a word that have 4 letters, and for each letter there are 9 possible choices, so the total outcome is 9*9*9*9=9^4.

Quote:
Hong in the 3 secretaries and 4 departments problem why di we get the solution as 3^4 and not 4^3? Please explain the concept

This is a very good question. In this question each secretary can have 0 to 4 reports to type, but each report must be and can only be typed by one person. Here you need to look which one is replaced back when you repeat the test. If a report is typed by one person, then it will not be typed by another person. In other words, the reports can not be replaced. On the other hand, if a secretary has typed one report, she'll still be placed in the pool for the second report. In other words, the secretary is the one that gets replaced when we repeat the test, and each report faces the same pool of three secretaries. Therefore, the outcome is 3*3*3*3 for the four reports.

Let me give you another example, see if it will help.

Example:
1. Five rooms are to be allocated to four people. How many ways are there if each room must be assigned to one person and one person only?
Here each people can have more than one room, or have no room at all. In other words the people will be replaced back into the pool in each repeated test and each room will face the same pool of people. So five room each face a pool of four people. 4*4*4*4*4=4^5

2. Five rooms are to be allocated to four people. How many ways are there if each person must be assigned to one room and one room only?
Here each room can be empty, or can have multiple people assigned to it. In other words the rooms will be placed back into the pool for each repeated test and each person will face same pool of five rooms. Therefore total outcome is 5*5*5*5=5^4.


Nice explanation.

Thanks,
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Re: Permutation, Combination and Probabilities [#permalink] New post 03 Sep 2009, 11:03
Quote:
Example:
There is a word AUSTRALIA. Four letters are taken at random. How many different words can be formed that have two vowels and two consonants?

Solution:
Vowels: A, U, A, I, A. Distinguish vowles: A, U, I.
Consonants: S, T, R, L

Outcomes with two vowels and two consonants:
1) The two vowels are different: C(3,2)*C(4,2)
2) The two vowels are the same: C(1,1)*C(4,2)
Total outcomes = 18+6=24



I dont understand the 2nd part for two vowels that are the same:2) The two vowels are the same: C(1,1)*C(4,2)
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Re: Tow Combination Problems [#permalink] New post 21 Sep 2009, 12:42
dvtohir wrote:
Guys, please help me with the following problems:

1. A committee of 3 people is to be chosen from four 4 married couples. What is the number of different committees that can be chosen if two people who are married to each other cannot both serve on the committee?

A.28 B.32 C.34 D.41 E.56

The answer to this question can be found at: gmatprep-practice-question-can-somebody-please-solve-it-78048.html
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Re: Permutation, Combination and Probabilities [#permalink] New post 19 Nov 2009, 17:34
Hi, I actually have a question on Binomial Distributions. How often do you see those (coin toss..etc.) problem in GMAT? From all practices and tests I did so far, I have not encountered any.
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Re: Permutation, Combination and Probabilities [#permalink] New post 19 Nov 2009, 17:48
Maybe once every 3-4 tests. Your question would be better answered by an instructor that takes the gmat yearly.
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Re: Permutation, Combination and Probabilities [#permalink] New post 21 Apr 2010, 20:31
great post!...thanks a ton hong hu :)
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Re: Permutation, Combination and Probabilities [#permalink] New post 07 Jun 2010, 12:58
Question: 3 prizes to 5 finalists out of 7 contestants.
I think it is: 7C5 x 5P3 = 21x60 = 1260.
Is this correct?
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Re: Permutation, Combination and Probabilities [#permalink] New post 22 Apr 2011, 10:40
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Sorry to resurrect this but is there a simpler way of answering the secretary question? (and is there an agreed answer, I've just trawled google and these boards to find several different solutions - including the 'OA' of 8/9 in the compilation available for download from this site :) )

Quote:
There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report?


My instinct is that the suggested answer in the file of 8/9 is far too close to 1, and starting by working out the probability that secretary A does not get a report to type <which leaves a number of other possibilities yet to remove> and subtracting from 1 gives: (2/3)^4 = 16/81 .... (which is greater than 1/9) .... 1-16/81 = 65/81

so P(everyone gets at least one) = 8/9 cannot be right.... can it? :-/


edit:: Having just put all 81 combinations into excel, there are clearly 36 ways to distribute the reports, 36/81 = 4/9) Now I just need to find the simple way of reaching that number......
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Re: Letters and Words [#permalink] New post 24 Apr 2011, 11:06
HongHu wrote:
AdrianG wrote:
HongHu wrote:
Letters and Words

A common GMAT P&C problem type is letter problems. You would be asked to take letters from an existing word to form new words.

Example:
Four letters are taken at random from the word AUSTRALIA. What is the probability to have two vowels and two consonants?

Solution:

Vowels: A, U, A, I, A.
Consonants: S, T, R, L

This is a case of "without replacement".
Total outcomes: C(9,4)=9*8*7*6/4!=126
Outcomes with two vowels and two consonants: C(5,2)*C(4,2)=60
Probability=60/126=10/21



Can anyone please explain why we can't count the probability like this?:

5/9 x 4/8 x 4/7 x 3/6 (VxVxCxC) = 5/63


The way you are doing it, you've given it a specific order.

For example, if there are 2 red balls and 2 white balls, what is the probability of getting 1 white ball and 1 red ball out in the first two draws? It's C(2,1)*C(2,1)/C(4,2) = 2/3.
You could write it out to verify:
R1R2
R1W1
R1W2
R2R1
R2W1
R2W2
W1R1
W1R2
W1W2
W2R1
W2R2
W2W1

But using your method you will have 2/4*2/3=1/3. Your problem is you have specified that you will draw a white ball first, and then red ball second. However there is also the possibility of drawing a red ball first, and then a white ball second. So you need to add another 1/3 to your result to get 2/3.



great post..i like your signature too..
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Re: Permutation, Combination and Probabilities [#permalink] New post 05 May 2011, 06:12
A fruit basket contains 1 apple, 2 bananas, and 4 cherries? If I pick four fruit at random without replacement, how many unique combinations are there?

Answer is easy to count, but I would like a formulaic solution.
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Re: Permutation, Combination and Probabilities [#permalink] New post 05 May 2011, 06:57
great collection.
will help in a big way.
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Re: Permutation, Combination and Probabilities   [#permalink] 05 May 2011, 06:57
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