Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 61396

Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 06:27
Question Stats:
54% (02:19) correct 46% (02:12) wrong based on 206 sessions
HideShow timer Statistics
Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b? (A) 5 (B) 6 (C) 11 (D) 16 (E) 19
Official Answer and Stats are available only to registered users. Register/ Login.
_________________



Manager
Joined: 27 Dec 2013
Posts: 190

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 06:43
my answer =19. The number of possible permutation without 2 in the first digit is 4X4X3X2X1=96 ways. The number of possible permuations with 5 in the second term is= 3X3X2X1=18 ways. 18c1/96c1 = 18/96= 3/16= > a+b=19. please post OA. Bunuel wrote: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5 (B) 6 (C) 11 (D) 16 (E) 19



Senior Manager
Joined: 07 Aug 2011
Posts: 492
Concentration: International Business, Technology

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 07:51
Bunuel wrote: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5 (B) 6 (C) 11 (D) 16 (E) 19 5!4!= 4*4! Perms not starting with 2 If 2nd element is 5 and 2 cannot come at 1st place then only 3,4 and 6 can take 1st place so total arrangements where 5 is 2nd element and 2 is not first are.....3*3! Probability = 3*3!/4*4! = 3/16 Answer 19



Manager
Joined: 23 May 2013
Posts: 180
Location: United States
Concentration: Technology, Healthcare
GMAT 1: 760 Q49 V45
GPA: 3.5

Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 09:54
Bunuel wrote: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5 (B) 6 (C) 11 (D) 16 (E) 19 The total number of permutations of this set is 5!. However, since the first digit cannot be 2, there are only 4 options for the first digit  the number of options for the rest of the digits remains the same. [4][4][3][2][1]. P(second digit is 5) = P(first digit is NOT 5)*P(second digit is 5) = (3/4)(1/4) = 3/16. Therefore a + b = 19. Answer: E



Retired Moderator
Joined: 06 Jul 2014
Posts: 1211
Location: Ukraine
Concentration: Entrepreneurship, Technology
GMAT 1: 660 Q48 V33 GMAT 2: 740 Q50 V40

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 13:23
Bunuel wrote: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5 (B) 6 (C) 11 (D) 16 (E) 19 At first we should calculate total variants first digit  all but 2  4 numbers possible second digit  all but 1 number is used  and 4 numbers possible third digit  all but 2 numbers is used  and 3 numbers possible fourth digit  all but 3 number is used  and 2 numbers possible fifth digit  all but 4 number is used  and 1 number possible product of all variants will equal to total variants 4 * 4 * 3 * 2 * 1 = 96 variants for calculating probability that the second term is 5 we should calculate number of variants with second term = 5 let's start from most restrictive part  second digit second digit  only 5 possible  1 variant first digit  all but 2  3 numbers possible third digit  all but 2 numbers is used  and 3 numbers possible fourth digit  all but 3 number is used  and 2 numbers possible fifth digit  all but 4 number is used  and 1 number possible product of all variants will equal to total variants with second term equal five: 1 * 3 * 3 * 2 * 1 = 18 variants so we have possibility \(\frac{18}{96} = \frac{3}{16}\) a + b = 3 + 16 = 19 Answer is E
_________________



Manager
Joined: 17 Mar 2015
Posts: 113

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
06 Apr 2015, 13:51
In order to satisfy our conditions we need to get first number as a non5 taking into account that 2 cannot happen either. Odds of that  3/4. Then we need to get 5 as second number provided only 4 numbers left: 1/4. Since these conditions are connected the resulting odds are 3/4*1/4 = 3/16, thus the answer to the question is 3+16 = 19. Option E.



Math Expert
Joined: 02 Sep 2009
Posts: 61396

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
13 Apr 2015, 06:43
Bunuel wrote: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for which the first term is not 2. A permutation is chosen randomly from S. The probability that the second term is 5 is given by a/b (in lowest terms). What is a+b?
(A) 5 (B) 6 (C) 11 (D) 16 (E) 19 VERITAS PREP OFFICIAL SOLUTION:Most Common Solution:What are the permutations of sequence S? They are the different ways in which we can arrange the elements of S. For example, 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc In how many different ways can we make the sequence? The first element can be chosen in 4 ways – one of 3, 4, 5 and 6. (You are given that 2 cannot be the first element). The second element can be chosen in 4 ways (2 and the leftover 3 numbers). The third element can be chosen in 3 ways. The fourth element can be chosen in 2 ways. And finally there will be only 1 element left for the last spot. Number of ways of making set S = 4*4*3*2*1 = 96 In how many of these sets will 5 be in the second spot? If 5 is reserved for the second spot, there are only 3 ways of filling the first spot (3 or 4 or 6). The second spot has to be taken by 5. The third element will be chosen in 3 ways (ignoring 5 and the first spot) The fourth element can be chosen in 2 ways. And finally there will be only 1 element left for the last spot. Number of favorable cases = 3*1*3*2*1 = 18 Required Probability = Favorable Cases/Total Cases = 18/96 = 3/16 = a/b a+b = 3 + 16 = 19 Answer (E) Intellectual Approach:Use a bit of logic of symmetry to solve this question without any calculations. Set S would include all such sequences as 3, 2, 4, 5, 6 or 4, 2, 3, 6, 5 or 6, 3, 4, 5, 2 etc – starting with 3, with 4, with 5 or with 6 with equal probability. By symmetry, note that 1/4th of them will start with 5 – which we need to ignore – so we are left with the rest of the 3/4th sequences. Now, in these 3/4th sequences which start with either 3 or 4 or 6, 5 could occupy any one of the 4 positions – second, third, fourth or fifth with equal probability. So we need 1/4th of these sequences i.e. only those sequences in which 5 is in the second spot. Probability that 5 is the second element of the sequence = (3/4)*(1/4) = 3/16 Therefore, a+b = 3+16 = 19 Answer (E)
_________________



Board of Directors
Joined: 17 Jul 2014
Posts: 2468
Location: United States (IL)
Concentration: Finance, Economics
GMAT 1: 650 Q49 V30
GPA: 3.92
WE: General Management (Transportation)

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
01 Jan 2016, 18:53
wow, now that's a tough son of a b... did everything correct until.. so we have 120 ways in which the numbers can be arranged, and 24 ways in which first digit is 2. thus, we have 96 ways in which 2 is not the first digit.
now to the second part: first digit  must be anything but not 2 or 5. so we have 3 possibilities second digit  must be 5, so 1 possibility. third digit  we are left with 3 possibilities fourth digit  2 possibilities first digit  1 possibility.
now here is where I made the mistake, instead of 3*1*3*2*1, I calculated 4*1*3*2*1.
instead of 3/16 and 3+16=19, I got 24/96 and 1/4 1+4=5.
damn me..I hate combinatorics/probability questions...



Intern
Joined: 12 Apr 2018
Posts: 13

Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
Show Tags
12 May 2019, 04:25
Sample event  96 possible permutations ( 4*4*3*2*1) Favorable event  18 possible permutations (3*1*3*2*1) a/b=18/96 OR 3/16 Therefore  a+b=3+16=19
Sent from my iPhone using GMAT Club Forum




Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi
[#permalink]
12 May 2019, 04:25






