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Math Expert V
Joined: 02 Sep 2009
Posts: 58368
Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi  [#permalink]

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15 00:00

Difficulty:   75% (hard)

Question Stats: 53% (02:17) correct 47% (02:12) wrong based on 202 sessions

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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

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Manager  Joined: 27 Dec 2013
Posts: 199
Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi  [#permalink]

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1

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.

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

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Kudos to you, for helping me with some KUDOS.
Director  Joined: 07 Aug 2011
Posts: 502
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27 Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi  [#permalink]

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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

Current Student B
Joined: 23 May 2013
Posts: 183
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]

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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.

.

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.

Retired Moderator Joined: 06 Jul 2014
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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]

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2
1
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
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Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi  [#permalink]

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In order to satisfy our conditions we need to get first number as a non-5 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 V
Joined: 02 Sep 2009
Posts: 58368
Re: Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi  [#permalink]

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1
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

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

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

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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  B
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]

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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, 05:25
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