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OG 195 197 Unsolved Mystery

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OG 195 197 Unsolved Mystery

by supermann » Fri Jun 13, 2008 10:06 pm
OG195

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Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length ?

(A) Six
(B) Eight
(C) Ten
(D) Fourteen
(E) Sixteen

-----------------------------------------------------------------------------------------------------------------

Do you guys have a simple mathematical shortcut? I solved it by actually drawing the line. That took me 5 minutes. I want to know the correct scientific way to do this. Many thanks.


OG197

Also,

If 75 percent of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percent answered both correctly?

Stuck on the Venn Diagram...Thanks.
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by Ian Stewart » Sat Jun 14, 2008 2:30 am
For the first problem, to go from X to Y in the shortest distance, your path must take 5 steps: you must go North three times, and East twice. So any path could be written as a 5-letter 'word' describing the steps we take in order. For example, if we go North the first 3 times, then East twice, we could write that as: NNNEE. Other paths would be NENEN, EENNN, etc. That's a long way of saying that this question is the same question as "How many 5-letter 'words' can you make using three Ns and two Es?" I'll assume that you've seen this kind of 'word counting' problem before- it's a basic problem type in counting, and any worthwhile GMAT prep book should explain how to do it- the answer is 5!/(2!*3!) = 10. If you haven't seen this before, let me know.
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by kindofbluenote » Sat Jun 14, 2008 5:14 am
Good explanation Ian, it took me a while to figure out that that was a question of basic combinations. Sometimes I think this test could ask me what 2+2 is, but ask in such a way that I wouldn't know how to answer it...

As for the second question, sometimes it's easier for me to use a formula than a Venn diagram for questions such as these. The one I use is (Category A + Category B + Neither - Both = Total.)

Since they're using percents, set the total students to 100. The formula then becomes 75+55+20-X=100. X=50.

Venn diagrams are useful for complicated problems, but sometimes they can make things more confusing (for me anyway, we all process things differently). Less complex problems that involve variables that fit neatly in the equation are better suited to using the equation method, I think.
There are three kinds of people in the world. Those who can count, and those who can't.
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by ildude02 » Sun Jun 15, 2008 3:52 pm
Ian Stewart wrote:For the first problem, to go from X to Y in the shortest distance, your path must take 5 steps: you must go North three times, and East twice. So any path could be written as a 5-letter 'word' describing the steps we take in order. For example, if we go North the first 3 times, then East twice, we could write that as: NNNEE. Other paths would be NENEN, EENNN, etc. That's a long way of saying that this question is the same question as "How many 5-letter 'words' can you make using three Ns and two Es?" I'll assume that you've seen this kind of 'word counting' problem before- it's a basic problem type in counting, and any worthwhile GMAT prep book should explain how to do it- the answer is 5!/(2!*3!) = 10. If you haven't seen this before, let me know.
Ian, can you please explain in the simplest way for the "How many 5 letter words can be formed from 3N's and 2E's", how did you come up with 5!/2! * 3!. I have found different explanations for these types of questions, so I'm still trying to see find the best possible way to approach these kind of problems. Appreciate it.
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by CredibleMonk » Sun Jun 15, 2008 5:35 pm
I am newcomer to this forum.

i read this post.

@ian ,your explanation is brilliant ...awesome.

For others,

General formula is: n!/(a!*b!*c!..) where a, b and c.. are the number of objects which are of the same kind.
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by Ian Stewart » Sun Jun 15, 2008 6:06 pm
ildude02 wrote:Ian, can you please explain in the simplest way for the "How many 5 letter words can be formed from 3N's and 2E's", how did you come up with 5!/2! * 3!.
I can give you three different ways to look at this, and hopefully one or more of them will be clear:

a) There is a formula you can learn, and it's sometimes quite useful. It's the formula CredibleMonk wrote out above. If you have a set of n letters, with a of one type, b of another, c of another, etc, then the number of n-letter words you can make is:

n!/(a!*b!*c!*...)

So, the number of 7-letter words you can make using each of these letters: A, A, A, B, B, C, C once each is 7!/(3!*2!*2!).

b) When you only have two different letters- by far the most common situation, both on the GMAT and in real life counting problems- you can look at it this way: say we have to make a five letter word from the letters N, N, N, E, E. We need to fill five slots _ _ _ _ _. Well, we just need to choose three slots for the three Ns; the Es will then have to go in the remaining two slots. Order isn't important, so the answer is 5C3 (or, you can choose 2 slots for the Es and get 5C2, which is equal to 5C3).

c) I prefer to think of it as follows: if all n letters are different, you can make n! words. But if I have, say, 3 letters that are the same, I need to divide by 3!, because I can rearrange these letters in 3! ways without changing the word I'm creating. And I'll need to do this for each repeated letter.

All of the above approaches will lead you to the same answer; I mention each, because particularly with counting problems, there are often many ways to look at each problem, and it's best to find a way that's most intuitive for you.

Finally, if you want to try a problem to put this type of counting into practice, in a GMAT-style question: If you flip a coin five times, what's the probability you get heads exactly twice?
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by ildude02 » Mon Jun 16, 2008 7:54 am
Ian, I appreciate your response. I didn't know about the formulae (a) until now. Inormally tend tyo follow the (b) option , which is more of the combinations and the permutations. I went with the combination approach and I used 5C3 * 5C2. My mistake was, I used 5C2 instead o 2C2 since there are 2 spots avialable and 2 E's to choose from. I hope I'm going the right direction with my approach.

As for the question about the coin flip/heads, which is a probability one, I see the total outcomes to be 2 ^5 and the favourable outcomes is (1/2) * (1/2). So the probability is 1/4 / (2 ^ 5) = 1/128. Let me know if I missed something.

Thanks again!
Ian Stewart wrote:
ildude02 wrote:Ian, can you please explain in the simplest way for the "How many 5 letter words can be formed from 3N's and 2E's", how did you come up with 5!/2! * 3!.
I can give you three different ways to look at this, and hopefully one or more of them will be clear:

a) There is a formula you can learn, and it's sometimes quite useful. It's the formula CredibleMonk wrote out above. If you have a set of n letters, with a of one type, b of another, c of another, etc, then the number of n-letter words you can make is:

n!/(a!*b!*c!*...)

So, the number of 7-letter words you can make using each of these letters: A, A, A, B, B, C, C once each is 7!/(3!*2!*2!).

b) When you only have two different letters- by far the most common situation, both on the GMAT and in real life counting problems- you can look at it this way: say we have to make a five letter word from the letters N, N, N, E, E. We need to fill five slots _ _ _ _ _. Well, we just need to choose three slots for the three Ns; the Es will then have to go in the remaining two slots. Order isn't important, so the answer is 5C3 (or, you can choose 2 slots for the Es and get 5C2, which is equal to 5C3).

c) I prefer to think of it as follows: if all n letters are different, you can make n! words. But if I have, say, 3 letters that are the same, I need to divide by 3!, because I can rearrange these letters in 3! ways without changing the word I'm creating. And I'll need to do this for each repeated letter.

All of the above approaches will lead you to the same answer; I mention each, because particularly with counting problems, there are often many ways to look at each problem, and it's best to find a way that's most intuitive for you.

Finally, if you want to try a problem to put this type of counting into practice, in a GMAT-style question: If you flip a coin five times, what's the probability you get heads exactly twice?
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by lokeshg98 » Mon Aug 30, 2010 11:16 am
A very important thing to note is that the grid structure is not really a symmetric one. ie counting the ways on the east and then doubling the number will not give the correct result.
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by saurabhmahajan » Mon Aug 30, 2010 11:27 pm
Hi Superman,

for OG 197...please solve using a grid as attached.
Hope that makes things simple for you.

I found answer to be [spoiler]50%[/spoiler] for OG 197
Attachments
OG 197.JPG
Thanks and regards,
Saurabh Mahajan

I can understand you not winning,but i will not forgive you for not trying.
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by diebeatsthegmat » Tue Aug 31, 2010 1:30 am
supermann wrote:OG195

Image

Pat will walk from intersection X to intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above. How many routes from X to Y can Pat take that have the minimum possible length ?

(A) Six
(B) Eight
(C) Ten
(D) Fourteen
(E) Sixteen

-----------------------------------------------------------------------------------------------------------------

Do you guys have a simple mathematical shortcut? I solved it by actually drawing the line. That took me 5 minutes. I want to know the correct scientific way to do this. Many thanks.


OG197

Also,

If 75 percent of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percent answered both correctly?

Stuck on the Venn Diagram...Thanks.
1. this is a combinatorical question.
you see that there is 4 ways ahead and 2 on the right thus matter which route the pers(on will go, she/he will travel 4 block ahead and 2 on the right,
this can be HHHHRR
= 6!/(4!*2!)= 10
answer is C
2, first of all, 100% students answered the question, no matter what they answered correctly or incorrectly
20% answered neither of the questions wrong correctly thus we have 100-20=80% answered both correctly and incorrectly
so the percentage we have to find= 75+55-80=50%
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