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Jacques wrote: ( Remember this question )
Judges will select 5 finalists from the 7 contestants entered in a singing competition. The judges will then rank the contestants and award prizes to the 3 highest ranked contestants: a blue ribbon for first place, a red ribbon for second place, and a yellow ribbon for third place. How many different arrangements of prizewinners are possible?
Can you please post the answer to this question?
Is it worked out as follows?
Ways to choose 5 finalists from 7 contestants: 7C5
Ways to chose the 3 highest rankers : 5C3*3!
hence total number of ways 7C5*5C3*3!



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This is a very good question. There had been much discussions on it if you search the forum.
The correct way to do this question is this: As we are only interested in the final three, the middle step is only throw in to confuse you. The question can be simplified as this: From 7 people we need to pick three people ranked. The answer is P(7,3)=7!/4!
The wrong answer is C(7,5)*P(5,3)=(7!/5!2!)*(5!/2!)=7!/2!2!
Since 2!*2!=4<4! it is obviouse that the wrong approach has a bigger answer. The reason is that when you do it in two steps there are inavoidably repeats in your answer. For example, compare these two cases: First, you choose A, B, C, D, E as the five people then you choose A, B, C as the final three. Second, you choose A, B, C, D, F as the five people and then you choose A, B, C as the final three. Obviously the final outcomes are exactly the same. However, you have counted it twice in the wrong approach. You can see that there is actually a very large number of repeats that are included in the wrong answer.
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Re: Letters and Words [#permalink]
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22 Sep 2005, 07:43
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
HongHu, this solusion is wrong, since it doesn't account for different orders of letters. This problem is actually more difficult.
Take for example just the first part  "Two vowels are different". It basicaly boils down to a question: how many four letter words with two vowels and two consonants can you build using three different vowels and four different consonants, taking them without replacement? There are much more than 18 such words.
Think about it, if we build words in order VVCC, than you already have 72 possibilities (P(3,2)xP(4,2)=72). What about diffent orders: CCVV, VCVC, CVCV? The total number of orders is 4!/(2!x2!)=6 (VVCC, CCVV, CVCV, VCVC, VCCV, CVVC). Each order takes 72 possibilites of arrangment of vowels and consonants, so the total # of words for point 1) is 6x72=432
Point 2)  "Two vowels are the same" You still have 6 ways to arrange VVCC, but only one way to use vowels (AA). The total number of words in point 2 is 6x1xP(4,2)=72.
So the total number of different 4 letter words with two vowels and two consonants that can be built using letters from AUSTRALIA is 504. (Additional practice: How many of these words actually exist in English?)
Do not underestimate the power of permutations...
BTW, I think that problems like this are way out of GMAT scope.



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KAPLAN COMBINATION QUESTION  PLEASE PROVIDE EXPLANATION [#permalink]
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07 Oct 2005, 14:37
If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5day period?
A  8/125
B  2/25
C 5/16
D  8/25
E  3/4



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FROM KAPLAN:
We will use the probability formula. The probability formula say that
Probability = Number of desired outcomes/Number of possible outcomes
First let's determine the number of possible outcomes of rain for City X over a 5day period. There are two possibilities for each day, rain or no rain, so the number of possible outcomes is 2 X 2 X 2 X 2 X 2 = 32.
Possibility of rain 3 out of 5 days = 5C3 =10
Thus, 10/32 = 5/16 =ANSWER C



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Re: ANSWER [#permalink]
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11 Oct 2005, 01:28
Tony wrote: FROM KAPLAN: We will use the probability formula. The probability formula say that Probability = Number of desired outcomes/Number of possible outcomes First let's determine the number of possible outcomes of rain for City X over a 5day period. There are two possibilities for each day, rain or no rain, so the number of possible outcomes is 2 X 2 X 2 X 2 X 2 = 32. Possibility of rain 3 out of 5 days = 5C3 =10 Thus, 10/32 = 5/16 =ANSWER C
You can use Binomial distribution for this one:
5C3 (1/2)^3 (1/2)^2 = 5/16



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One word of caution for using the binomial distribution. The events need to be independent.
If you're working a dependent event, use the hypergeometric distrubution instead.



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Re: Letters and Words [#permalink]
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12 Oct 2005, 12:53
JTB wrote: 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 HongHu, this solusion is wrong, since it doesn't account for different orders of letters. This problem is actually more difficult. Take for example just the first part  "Two vowels are different". It basicaly boils down to a question: how many four letter words with two vowels and two consonants can you build using three different vowels and four different consonants, taking them without replacement? There are much more than 18 such words. Think about it, if we build words in order VVCC, than you already have 72 possibilities (P(3,2)xP(4,2)=72). What about diffent orders: CCVV, VCVC, CVCV? The total number of orders is 4!/(2!x2!)=6 (VVCC, CCVV, CVCV, VCVC, VCCV, CVVC). Each order takes 72 possibilites of arrangment of vowels and consonants, so the total # of words for point 1) is 6x72=432 Point 2)  "Two vowels are the same" You still have 6 ways to arrange VVCC, but only one way to use vowels (AA). The total number of words in point 2 is 6x1xP(4,2)=72. So the total number of different 4 letter words with two vowels and two consonants that can be built using letters from AUSTRALIA is 504. (Additional practice: How many of these words actually exist in English?) Do not underestimate the power of permutations... BTW, I think that problems like this are way out of GMAT scope.
You are very right, JTB. Thanks for point that out. After you've got the four letters you definitely need to order them to get different outcomes. So the first case (with different vowels) would be C(3,2)*C(4,2)* P(4,4)=432, and the second case (with same vowels) would be C(1,1)*C(4,2)* P(4,4)/2=72, thus they add up to 504.
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I also agree with you that this question would be in the high end of the GMAT scope. However if you follow the rules and know how to think it shouldn't be very hard to get to the answer.
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HongHu wrote: Very good question. When we do 9^4, 5^2 and 4^2 we've already taken count for the cases of different orders. For example, for the two vowels AU and UA are counted as two occasions. We don't need to order those four letters again. In other words we don't need to multiply 5^2*4^2 by 4!. However, what we have not counted, is that the two vowels can be placed onto different space. For example, for each set of two vowels and two consonants, say AU and ST, we could do different placements: AUST, ASUT, or ASTU, etc. They are all different words that need to be counted. Therefore we need to pick two spaces for the two vowels out of the four spaces to account for this. That's why we multiply 5^2*4^2 by C(4,2). The rationale is the same with the binomial distribution formula. On looking back I noticed that I have mistakenly used P(4,2) in the solution. It is wrong. We've already counted different orders, so we need combination instead of permutation. Thanks for catching that.
I am having trouble following the reasoning here. If we all agree that different words can be formed by picking a different order, it sounds like 4 letters can be arranged among themselves in 4! ways, not C(4,2).



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C(4,2) is not really dealing with the ordering. It is the number of ways you can pick two positions out of four for the two vowels. We don't need to order them because in the solution they are already ordered. As the example you quoted in last post, AU and UA are counted as 2 different outcomes when you do 4^2, so you don't need to order them again.
For example, if I have two letters, A and B, what is the total number of two letter words if we can use a letter repetitively? We have 2^2. The first position can have A or B, and the second postion also can have A or B. So total outcome is 4. You don't need to multiply this result by 2! because we've already counted all the different orders.
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Re: Repeated Outcomes [#permalink]
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25 Nov 2005, 13:48
HongHu wrote: Repeated Outcomes
If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.
Explanation: For each object, there are k outcomes. So the total number of outcomes would be k*k*k ...*k, for n times.
Example:
Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there are some fruits in the basket, how many different choices does she have? (Assuming the fruits are all different from each other.) Total number of fruits = 9 For each fruit there are 2 possible outcomes: included, not included Total outcome = 2^91 (The minus one is to take out the one possible outcome where nothing is in the basket.)
Example:
Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.) Total outcome = (2^31)*(2^41)*(2^21)
Example:
There are three secretaries who work for three departments. If each of the three departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^3=27 Total outcome that three secretary are assigned at least one report: P(3,3)=3!=6 Probability = 6/27=2/9
Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36 Probability = 36/81=4/9
Question: in the above example, why isn't total outcome = 4^3?
Could you plz, answer this question. I'm a little confused
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Re: Repeated Outcomes [#permalink]
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26 Nov 2005, 15:51
rlevochkin wrote: HongHu wrote: Repeated Outcomes
If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n.
Explanation: For each object, there are k outcomes. So the total number of outcomes would be k*k*k ...*k, for n times.
Example:
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81
Question: in the above example, why isn't total outcome = 4^3?Could you plz, answer this question. I'm a little confused
Yes this is always the part that confuses me the most too. The key is you need to make sure which is k and which is n. The object would be the ones that must have one and only one potential item assigned to it. In this example, each report must be typed by a secretary, and can't be typed twice. So report is the object, each with three possible outcomes. In other words, your total outcome would be 3*3*3*3=3^4. In this case a secretary can type multiple reports. It is even possible that secretary A has to type all four reports and all the others don't have to do anything (AAAA).
If the question is changed to this: Three secretaries each must choose one of the four reports to type (say it is training materials and it doesn't matter if one report is typed multiple times), then secretaries would be the object who needs to assigned one and only one potential outcome. Total outcome in this case would be 4*4*4=4^3. Here it is possible that all three secretaries type the same report (222, for instance).
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Need help clarifying... [#permalink]
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03 Dec 2005, 22:19
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
HongHu, I wonder why the repeating letters don't matter here. Consider this example.
There are 3 letters, A, B, C. If we select 2 letters out of these 3, there will be 3C2 arrangements = 3 which are AB, AC and BC
However, if 3 letters are A, B, B (two letters are repeated). If we select 2 letters out of 3, there will be only 2 arrangements which are AB and BB.
So that shows the repeating letters matter?
I would appreciate it if you can help clarify this....Thanks!



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Re: Need help clarifying... [#permalink]
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19 Dec 2005, 07:55
gmatbangkok wrote: 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
HongHu, I wonder why the repeating letters don't matter here. Consider this example. There are 3 letters, A, B, C. If we select 2 letters out of these 3, there will be 3C2 arrangements = 3 which are AB, AC and BC However, if 3 letters are A, B, B (two letters are repeated). If we select 2 letters out of 3, there will be only 2 arrangements which are AB and BB. So that shows the repeating letters matter? I would appreciate it if you can help clarify this....Thanks!
I believe you might have to consider the two B(s) as two entities, say, one is B1 and the other is B2. So AB1, AB2 would mean different things.



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Example: To find all permutations on A B C C
Answer: ABCC, ACBC, ACCB,
BACC, BCAC, BCCA,
CABC, CBAC, CBCA,
CACB, CCAB, CCBA
There are 12 orders, or 12 permutations in total.
P( permutation of n objects with n1 identical ones)
= n!/ n1!
Example of ABCC:
n = 4; (A,B,C,C)
n1= 2; (C,C)
n!/ n1! = 4! / 2! = (4 x 3 x 2 x 1)/ (2 x 1) = 12



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Back to the topic Repeated Outcomes:
Code: If there're only k possible outcomes for each object, total possible outcomes for n objects is k^n. and example: Code: Ann is making a fruit basket with 3 apples, 4 plums and 2 grapefruits. If she can include however many fruits as she wants in the basket as long as there is at least one of each kind, how many different choices does she have? (Assuming the fruits are all different from each other.) Total outcome = (2^31)*(2^41)*(2^21)
Let's take 4 plums. If I have to choose atleast 1 plum, then the outcome is
C(4,1)+C(4,2)+C(4,3)+C(4,4) = 4+6+6+1 = 17.
where as 2^41=15.
Could somebody explain to me know why my answer/approach is wrong.
I am sorry, if I am interrupting the flow of messages or posting in wrong location.
Thanks



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C(4,3)=4, instead of 6.
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Re: Repeated Outcomes [#permalink]
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02 Aug 2006, 17:29
HongHu wrote: Repeated Outcomes
There are three secretaries who work for four departments. If each of the four departments have one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretary are assigned at least one report?
For each report there are three possible outcomes: Secretary A, Secretary B, Secretary C Total outcome: 3^4=81 Total outcome that three secretary are assigned at least one report: We need to choose one secretary C(3,1) to be assigned of two reports C(4,2) and then the rest two secretary each to be assigned of one report P(2,2). C(3,1)*C(4,2)*P(2,2)=36 Probability = 36/81=4/9
Hi everyone,
Iâ€™m a newbie posting for the first time. First off â€“ great thanks for all the material here  premium stuff. Hope youâ€™ll help me defeat my logic.
This question drives me mad  I understand why the total number of outcomes is 81 and why the total number of outcomes when each secretary gets at least one report is 36, but when I am trying to use logic instead of formula, the former seems to be giving up. Please let me know where I am wrong.
So, my understanding is that the total number of outcomes when 4 reports (A,B,C,D) are assigned randomly among 3 secretaries (1st, 2nd, 3rd) is as follows:
1) all 4 reports are assigned to only one secretary  3 outcomes, plus
2) 3 reports out of four (4C3) are assigned to one secretary (*3) and the fourth is assigned to either secretary (*2) => 4C3*3*2 = 4*3*2=24 outcomes, plus
3)  two reports out of four (4C2) are assigned to a secretary (*3) and the other two are assigned like this: both to one secretary, both to the other secretary, one report per each secretary and vice versa, i.e. 4 options > (*4).
To make the last one clearer, let's say when reports AB go to 1st sec., there are 4 outcomes: first â€“ reports CD both go to 2nd sec.; second  CD go both to 3rd sec.; third  report C goes to 2nd sec. and D to 3rd sec.; and the fourth outcome  C goes to 3rd sec. and D goes to 2nd sec. So, 4C2*3*4 = 6*3*4 = 72
Thus, we have 3+24+72 = way out of 81. I understand I have repetitive counts, but can't figure out where.
Thanks in advance.



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Re: Letters and Words [#permalink]
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03 Aug 2006, 08:10
HongHu wrote: Letters and Words
Vowels: A, U, A, I, A. Consonants: S, T, R, L
Example: From the word AUSTRALIA, a letter is taken at random and put back. Then a second letter is taken and put back. This process is repeated for four letters. What would be the possibility that two vowels and two consonants have been chosen?
This is a case of "with replacement": Total outcomes: 9^4 Outcomes with two vowels and two consonants: C(4,2)*5^2*4^2 You can calculate the probability from here.
Shouldn't this be just 5/9*5/9*4/9*4/9?




Re: Letters and Words
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