Permutations, Combinations, Probability - Download Questions : GMAT Quantitative Section - Page 4
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06 Jan 2011, 05:31
philiptraum wrote:

Quick question on number 4 though:

How many 3-digit numbers satisfy the following conditions: The first digit is different from zero and the other digits are all different from each other?

From my reading of this "the other digits" implies the 2nd and 3rd digits and not the first. Well if only the 2nd and 3rd digits are different from each other then shouldn't it be 9*10*9?

I agree with you on that one.
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11 Jan 2011, 12:51
thanks. great stuff.
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29 Jan 2011, 07:23
You've heard this 74 times, but it's well worth the 75th.

So, thanks!!! It so happens that I was beginning to wonder how I was going to practice this topic after learning it from the web (considering it is a weak area) and I found this.

Thanks again!
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14 Feb 2011, 01:20
Thanks for sharing!
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15 Feb 2011, 10:05
Thank you! Appreciate it very much!
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17 Apr 2011, 01:17
Thanks, it was really helpful !
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23 Apr 2011, 01:56
Awesome resource Just one small issue with the secretaries/reports answer, I think there's an error with the OA... 8/9 is clearly too big to be correct

 "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? "

Spoiler:
[Reveal] Spoiler:
There are 3^4 (or 3*3*3*3) possible combinations = each option therefore is 1/81.

You then choose 1 of three secretaries (3C1) to receive two reports (4C2), and then work out the number of permutations to assign the remaining 2 reports to the 2 remaining secretaries (2P2).

3C1 * 4C2 * 2P2 = 36

[Reveal] Spoiler:
= 36/81 = 4/9

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28 Apr 2011, 08:04
In the second file, the solution for Q32 is wrong. The correct answer should be 171.
The total number of diagonals is 21x18/2=189 and from this, one has to remove the number of diagonals from one vertex, i.e. 189-18=171. To the two adjacent vertices to the one which doesn't send any, there are still 18 diagonals connected to. When working out the number of the diagonals one should take into account the definition of a diagonal - a line segment which connects two non-adjacent vertices. So, any vertex, connects to n-3 different vertices (not to itself, and not to the two adjacent vertices).

Question 33 is not properly formulated, because the answer depends on how the three vertices relate to each other. Of course the solution is wrong, due to the same mistake made for the previous question. So, one should ask the question how many diagonals are in a polygon with 18 vertices in which three adjacent vertices don't send any. And the correct answer is 91.
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12 Aug 2011, 23:07
prasrush2001 wrote:
don't have words for you to thank..

You are very welcome!
Great collection of questions, very useful!
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09 Sep 2011, 22:54
Some of the Answers are wrong in this document.
Heres One

20. In a flower shop, there are 5 different types of flowers. Two of the flowers are blue, two are red and one is yellow. In how many different combinations of different colors can a 3-flower garland be made?

a) 4.
b) 20.
c) 3.
d) 5.
e) 6.

According to the document

20. The best answer is A.
We want to make a 3-flower garlands, each should have three colors of flowers in it.
There are two different types of blue and two different types of red.
The options are (2 blue) x (2 red) x (1 yellow) = 4 options.

According to me

because it according the the question it is not compulsory to have 3 colors ! it asks for a number of different combinations i see the following as combination

1) 2Red + 1 Blue
2) 2red+1 yellow
3)1 yellow+1 blue+ 1 red
4) 2 blue+ 1 yellow
5) 2 blue + 1 red

thats a total of 5
what do u say
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09 Sep 2011, 23:57
26. What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8

26. The best answer is D.
There are three different arrangements of a boy and two girls:(boy, girl, girl), (girl, boy, girl), (girl, girl, boy). Each has a probability of (1/2)3. The total is 3*(1/2)3=3/8.

WHY DOES THE ORDER MATTER IN THIS QUESTION. I MEAN THEY HAVE A BOY AND 2 GIRLS SHOULD NOT IT JUST BE 3/8?
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09 Oct 2011, 17:48
thank you
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03 Dec 2011, 10:30
can any send tat link plz
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27 Dec 2011, 21:41
The answer to question 40 does not seem right to me. Shouldn't the answer be 48? The question is as follows.
A computer game has five difficulty levels. In each level you can choose among four different scenarios except for the first level, where you can choose among three scenarios only. How many different games are possible?
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19 Jan 2012, 19:57
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27 Mar 2012, 06:34
Really looking forward to this set. Thanks mate.
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27 Mar 2012, 11:18
cmal1985 wrote:
26. What is the probability for a family with three children to have a boy and two girls (assuming the probability of having a boy or a girl is equal)?

a) 1/8
b) ¼
c) ½
d) 3/8
e) 5/8

26. The best answer is D.
There are three different arrangements of a boy and two girls:(boy, girl, girl), (girl, boy, girl), (girl, girl, boy). Each has a probability of (1/2)3. The total is 3*(1/2)3=3/8.

WHY DOES THE ORDER MATTER IN THIS QUESTION. I MEAN THEY HAVE A BOY AND 2 GIRLS SHOULD NOT IT JUST BE 3/8?

What about the possibility of 3 boys or 3 girls ? Here, I think the order does not matter.
the possibilities are - 3Boys, 2Boys-1Girl, 1boy-2girls, 3Girls. so out of 4 possible outcomes, the probability of 1boy-2girls is 1/4.

What is wrong with this reasoning ?
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12 Sep 2012, 18:52
Thank you so much! Keep up the good work~
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18 Jan 2013, 07:10
Looks like there some wrong solutions as pointed out by other members , has the doc been updated to reflect this ?just checking
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16 Feb 2014, 07:02
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