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Probability and Combination Problems

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GMAT/MBA Expert

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Probability and Combination Problems

Mon Oct 22, 2007 1:04 pm
Found this great set of probability and combination problems. Remember--these types of questions appear infrequently on the GMAT, so don't over-emphasize them too much in your studies!
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K23 Newbie | Next Rank: 10 Posts
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Mon Oct 11, 2010 8:53 am
can anybody explain the problem 27) in how many ways can you sit 8 people on the bench if 3 of them must sit together?
thanks.

eladoren Newbie | Next Rank: 10 Posts
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Wed Oct 06, 2010 12:34 pm
regarding question #32
The answer 170 is not correct in my opinion. it should be 171
the suggested solution is not considering the diagonal between the two neighbors of the un-attached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)-18=171
any opinions ?

thunderdogg Junior | Next Rank: 30 Posts
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Mon Nov 26, 2007 8:14 pm
Question 6 from this problem set:

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.

---
So um... how does a 5 digit credit card have a 6th digit that is used to determine the 5th digit. This problem set might be a waste of time. ONly on problem 6 and the questions are clearly not very good.

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Mon Nov 26, 2007 9:25 pm
Thanks for this feedback, thunderdogg. Anyone else have opinions as to whether this problem set is useful?

If the consensus seems to be that it sucks then I'll take it down.

samirpandeyit62 Master | Next Rank: 500 Posts
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Mon Nov 26, 2007 9:32 pm
Hi Eric,
I think its a good set for P&C coz it has so many different types of P&C problems, so it would be really good for practise & to learn the general approach to solve diff kinds of P&C probs, I've not seen each & every soln though, some may be wrong, but the same can be discussed.

Thanks for posting the doc.

_________________
Regards
Samir

Tue Nov 27, 2007 12:05 pm
this is thunderdogg, the set is overall pretty decent.

it just gets off to a rocky start. with the 1st and 2nd questions being identical and then question 6 (as i've already shown). But it gets pretty good. its worth keeping up, sorry for judging it to quickly.

tripathimani Junior | Next Rank: 30 Posts
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Fri Dec 26, 2008 4:09 pm
Thanks Eric for putting this up.

It has good variety of questions and it gains momentum after question 28.

UCLA97grad Newbie | Next Rank: 10 Posts
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Mon Sep 07, 2009 1:44 pm
This is an excellent resource.

I also found these problems on Scribd to be extremely difficult.
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Alex Lbn Newbie | Next Rank: 10 Posts
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Wed Jul 28, 2010 9:28 am
Original post: thunderdogg
Posted: Mon Nov 26, 2007 10:14 pm
Question 6 from this problem set:

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.

---
So um... how does a 5 digit credit card have a 6th digit that is used to determine the 5th digit. This problem set might be a waste of time. ONly on problem 6 and the questions are clearly not very good.

This problem actually makes sense. We have 5 slots which we need to fill with different numbers. And we are given that the first and the second are 1 & 2 respectively.
_1_ _2_ __ __ __

The third digit is bigger than 6, therefore it could be either 7 or 8 or 9, since the range is [1:9].
The fourth digit is divisible by 3, therefore it could be either 3 or 6 or 9.
The fifth digit is 3 times the sixth, therefore it's multiple of 3, so in a given range it could be either 3 or 6 or 9. By the way, the sixth digit could be either 1 or 2 or 3, since the product of 5th and 6th can't be more than 9. So the reference to the 6th digit in the question is absolutely sensible.
From this point, it looks like a simple combinatorics problem. We have 3 options in each of three decisions. So to calculate possible numbers we need to multiply 3 by 3 by 3. So the answer is 27 (A).

boysangur Junior | Next Rank: 30 Posts
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Sun Aug 15, 2010 12:25 pm
Thanks a LOT for posting this. I'm going through Kaplan right now and these types of problems are kicking my butt. I could never fully wrap my head around these problems, I don't know why.

More on topic: has anyone looked at #24? In the answers, it says that the probability of choosing an even number is 1/2, when we have 8 digits available (0 through 9, without 1 or 4). How is the probability 1/2 though? Aren't there 3 even numbers (2, 6 and 8) and 4 odd numbers (3, 5, 7, and 9)? Are they considering zero to be an even number? Both Kaplan and Princeton Review are claiming it isn't. Is the test wrong? Are the books? Am I crazy?

K23 Newbie | Next Rank: 10 Posts
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Mon Oct 11, 2010 8:54 am
This is an excellent resource.

I also found these problems on Scribd to be extremely difficult.
hi,

ga1ex Newbie | Next Rank: 10 Posts
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Fri Aug 17, 2012 10:39 pm
This problem set is super helpful - thanks for posting.

I think I've found some more incorrect answers:

#37. The answer doesn't take into account the possibility of the second and third digits being the same. I think the right answer is C.

#41. The answer takes into account the possibility of the first digit being 0, and excludes that as an option, but it doesn't do the same for the possibility of the first and second digits being 0, or first second and third, or all four. I'm not sure what the right answer is, but I think it's somewhere around 3515.

Am I missing something?

ga1ex Newbie | Next Rank: 10 Posts
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Fri Aug 17, 2012 11:35 pm
regarding question #32
The answer 170 is not correct in my opinion. it should be 171
the suggested solution is not considering the diagonal between the two neighbors of the un-attached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)-18=171
any opinions ?
I think the equation they use in the solution actually does take that diagonal into account. Try it with a pentagon or hexagon.

ga1ex Newbie | Next Rank: 10 Posts
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Sun Aug 19, 2012 9:17 pm
Can anyone help me with #47? I understand why the answer key's answer is correct, but my approach was to figure out the probability that one secretary gets no reports (16/27) and the probability that two secretaries get no reports (1/27) and subtract those probabilities from 1, and the result is different (10/27). Can someone please explain why my approach doesn't work?

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