Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 13 Dec 2006
Posts: 297
Location: Indonesia

Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
Updated on: 26 Apr 2018, 13:11
Guys, Enclosed please find list of 55 different question with explanation. I got this from the GMATClub forum and found very impressive to learn most of the required commonly tested concepts. I am returning back to the forum so that everyone can use it for hir/her benefit. Amar Please discuss specific questions in PS or DS subforums.
MODERATOR
Originally posted by Amardeep Sharma on 15 Dec 2007, 17:05.
Last edited by Bunuel on 26 Apr 2018, 13:11, edited 1 time in total.
UPDATED.




Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
24 Oct 2009, 07:56




Manager
Joined: 05 Jul 2009
Posts: 124

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
25 Oct 2009, 20:03
I have a question regarding the #37 question and the answer given.
37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?
a) 2520 b) 3150 c) 3360 d) 6000 e) 7500
37. The best answer is A. The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (103 used before) and the fifth has 6 options (104 used before). The total is 4*5*3*7*6=2520.
Now, it is not mentioned that a number cannot be repeated in for the 3rd digit. Say in the 2nd digit the number is 3 and so it is for the 3rd digit. The structure can be 233, now for the 4th digit the option is 8 not 7.
Surely then the answer given is not right. But I don't know how to proceed further. Can someone please help? Or am I wrong about the porcess?



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
25 Oct 2009, 21:12
eresh wrote: I have a question regarding the #37 question and the answer given.
37. How many 5 digit numbers can be created if the following terms apply: the leftmost digit is even, the second is odd, the third is a non even prime and the fourth and fifth are two random digits not used before in the number?
a) 2520 b) 3150 c) 3360 d) 6000 e) 7500
37. The best answer is A. The first digit has 4 options (2,4,6,8 and not 0), the second has 5 options (1,3,5,7,9) the third has 3 options (3,5,7 and not 2), the fourth has 7 options (103 used before) and the fifth has 6 options (104 used before). The total is 4*5*3*7*6=2520.
Now, it is not mentioned that a number cannot be repeated in for the 3rd digit. Say in the 2nd digit the number is 3 and so it is for the 3rd digit. The structure can be 233, now for the 4th digit the option is 8 not 7.
Surely then the answer given is not right. But I don't know how to proceed further. Can someone please help? Or am I wrong about the porcess? You are absolutely right. I'd say more there is no correct answer in choices. Surely the second and the third digits can be the same (in 3 cases out of 15) and in this case the number of possibilities for the 4th and 5th will be 8 and 7 and not 7 and 6. So here is my solution: 1. second and third digits are the same: 4*3*1*8*7 2. second and third digits are not the same: 4*4*3*7*6 Total=4*3*1*8*7+4*4*3*7*6=2688 P.S. eresh, can you please post other questions (you consider as worth of discussion) in PS or DS forums, as not many check this tree for particular problems, but rather for the tips and tutorials. You may even duplicate this question as there could be other opinions about it. Thanks.
_________________



Manager
Joined: 05 Jul 2009
Posts: 124

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
25 Oct 2009, 21:46
Bunuel, I believe your solution is absolutely correct.
So here is my solution: 1. second and third digits are the same: 4*3*1*8*7
2. second and third digits are not the same: 4*4*3*7*6
Thinking of the approach, I understand that there are a total 15 combination of the 2nd and 3rd digit, of which in 3 combination the digits will be same. Therefore, there are 12 other possibilities where the digits are not same. The red part 4*3 also is equal to 12 but I could not figure out why it was 4*3 (My brains seems to stop working when I try to focus too much :D). Anyway, your answer seems correct and thank you for that.



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
25 Oct 2009, 22:26
eresh wrote: Bunuel, I believe your solution is absolutely correct.
So here is my solution: 1. second and third digits are the same: 4*3*1*8*7
2. second and third digits are not the same: 4*4*3*7*6
Thinking of the approach, I understand that there are a total 15 combination of the 2nd and 3rd digit, of which in 3 combination the digits will be same. Therefore, there are 12 other possibilities where the digits are not same. The red part 4*3 also is equal to 12 but I could not figure out why it was 4*3 (My brains seems to stop working when I try to focus too much :D). Anyway, your answer seems correct and thank you for that. Think about it this way: we don't want second and third digits to be the same, let's first choose the third digit(clearly it doesn't matter which one we choose first) how many possibilities are there? 3 (3 odd primes), than choose the second digit how many possibilities are there as one odd prime is already used? 51=4 > 3*4=4*3. Hope now it's clear.
_________________



Manager
Joined: 30 Nov 2010
Posts: 198
Schools: UC Berkley, UCLA

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
Updated on: 08 Dec 2010, 07:52
25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? a) 15/16 b) 11/16 c) 11/12 d) ½ e) 5/8 29. In a jar there are 15 white balls, 25 red balls, 10 blue balls and 20 green balls. How many balls must be taken out in order to make sure we took out 8 of the same color? a) 8 b) 23 c) 29 d) 32 e) 53 30. In a jar there are 21 white balls, 24 green balls and 32 blue balls. How many balls must be taken out in order to make sure we have 23 balls of the same color? a) 23 b) 46 c) 57 d) 66 e) 67 I don't quite understand the explanations to these problems. Can someone please help me out? Thanks in advance
Originally posted by mariyea on 08 Dec 2010, 07:15.
Last edited by mariyea on 08 Dec 2010, 07:52, edited 1 time in total.



Math Expert
Joined: 02 Sep 2009
Posts: 64951

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
08 Dec 2010, 07:21
mariyea wrote: 25. John wrote a phone number on a note that was later lost. John can remember that the number had 7 digits, the digit 1 appeared in the last three places and 0 did not appear at all. What is the probability that the phone number contains at least two prime digits? a) 15/16 b) 11/16 c) 11/12 d) ½ e) 5/8 I don't quite understand the explanation to this problem. Can someone please help me out? Thanks in advance First of all: As for the question: The phone numbers is of a type: {X}{X}{X}{X}{1}{1}{1}. {X}'s can take following values: 4 primes {2, 3, 5, 7} and 4 nonprimes {4, 6, 8, 9}. Total 8 choices for each {X}. Probability that {X} will be prime is therefore \(\frac{4}{8}=\frac{1}{2}\) and probability of {X} will not be a prime is again \(\frac{1}{2}\). We want at least 2 {X}'s out of 4 to be primes, which means 2, 3 or 4 primes. Let's count the opposite probability and subtract it from 1. Opposite probability of at least 2 primes is 0 or 1 prime: So {P}{NP}{NP}{NP} and {NP}{NP}{NP}{NP}. Scenario 1 prime  {P}{NP}{NP}{NP}: \(\frac{4!}{3!}*\frac{1}{2}*(\frac{1}{2})^3=\frac{4}{16}\). We are multiplying by \(\frac{4!}{3!}\) as scenario {P}{NP}{NP}{NP} can occur in several different ways: {P}{NP}{NP}{NP}, {NP}{P}{NP}{NP}, {NP}{NP}{P}{NP}, {NP}{NP}{NP}{P}  4 ways (basically the # of permutations of 4 objects out ow which 3 are the same). Scenario 0 prime  {NP}{NP}{NP}{NP}: \((\frac{1}{2})^4=\frac{1}{16}\). Hence opposite probability = \(\frac{4}{16}+\frac{1}{16}=\frac{5}{16}\). So probability of at least 2 primes is: 1(Opposite probability) = \(1\frac{5}{16}=\frac{11}{16}\) Answer: A.
_________________



Intern
Joined: 11 Apr 2011
Posts: 20
Location: London
WE 1: 3 yrs f/t manager in aerospace (IT  Commercial  Int'l Mktg)
WE 2: 5 yrs freelance consultant to Gov't

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
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: 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 Answer:



Director
Joined: 22 Mar 2011
Posts: 576
WE: Science (Education)

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
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. 18918=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 nonadjacent vertices. So, any vertex, connects to n3 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.
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Intern
Joined: 24 Nov 2018
Posts: 7

Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
Show Tags
05 Sep 2019, 07:48
I have got A) 2520 as answer. 1 assumption I have used is no digits can be repeated.
Consider a 5 digit number _ _ _ _ _ Left most  Possibility : 2,4,6,8
2nd Digit Odd  (As I have mentioned that no digit is repeated) No. of ways 4 numbers can be selected out of 5 odd numbers = 5C4 = 5
3rd Digit  Only 3,5,7 are possible here.
4th Digit  (103)C1 = 7C1
5th Digit  (104)C1 = 6C1
So total possible such numbers = 4 * 5 * 3 * 7 * 6 = 2520
Please point out any flaw in the solution.
Posted from my mobile device




Re: Permutations, Combinations, Probability  Download Questions
[#permalink]
05 Sep 2019, 07:48




