Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi

Question

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

(A) 5
(B) 6
(C) 11
(D) 16
(E) 19

GSBae · Accepted Answer

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.

Answer: E

Harley1980 · Answer

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

shriramvelamuri · Answer

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.

Lucky2783 · Answer

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

Zhenek · Answer

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.

Bunuel · Answer

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)

mvictor · Answer

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...

sanjayparihar16 · Answer

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

Getsome2 · Answer

Bunuel   VeritasKarishma

Trying to wrap my head around this logic of symmetry approach. Is my logic as set forth below correct from a symmetry perspective?

Based on the question, #2 cannot start the sequence, thus we have four TOTAL options (#3,#4,#5 and #6) for Slot One
  Since, we're trying to find the probability of #5 appearing in Slot Two, then we need to remove #5 from our options as outcomes for Slot One
So now we have 3/4 - i.e.  /

For Slot Two now, we've 'used' one digit already, however, we've also 'gained' an option back in the form of digit #2 - so our TOTAL options are four (#2 plus the three digits that were not selected in Slot One)
  The desired outcome we're looking for is digit #5 - thus we have one option
So now we have 1/4 - i.e.  /

So now, in order for us to get to Slot Two, we must draw a digit for Slot One FIRST - which is why we must consider the probability of Slot One and multiply the 1/4 by the 3/4 in order to get 3/16 as the probability for #5 appearing in Slot Two

I'm not sure if the above applies the symmetry principle or simply another way to approach the problem?
According to the above, then in a sense, it also seems like we don't care about what happens in Slots Three, Four and Five?

Thank you for your help and explanations on the above as well as broadly speaking.

EgmatQuantExpert · Answer

Solution 
 Given

• 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 and the probability that the second term is 5 is given by  \frac{a}{b } (in lowest terms).

To Find

• The value of a+ b.

Approach and Working Out

• Total number of permutations = 4 × 4 × 3 × 2 × 1.
 o The first digit cannot be 2.
o After that 4 places and 4 digits. 
• Total number of permutation when 5 is the second digit is = 3 × 1 × 3 × 2 × 1.
 o The first 3 as we cannot put 2 or 5 here.
o 2nd digit must be 5.
o After that 3 places and 3 digits. 
• Required probability,  \frac{a}{b}  =  \frac{(3 × 1 × 3 × 2 × 1)}{( 4 × 4 × 3 × 2 × 1)} 
 o  \frac{a}{b}  = \frac{3}{16} 
o a + b = 19

Correct Answer: Option E

KarishmaB · Answer

Yes, you are right. 
There are 4 options for first digit (3/4/5/6) in S but we are interested in instances that have only 3 of those 4 options (3/4/6).
So probability of picking first digit this way = 3/4

There are 4 options for second digit (2/leftover 3 options) but we are interested in only 1 of those (5)
So probability of picking second digit this way = 1/4

After that, anything goes.

So (3/4)*(1/4) = 3/16

pierjoejoe · Answer

can anyone check this?

2,3,4,5,6
for the first digit we can either pick {5} or one of {3,4,6}.
possible first digits {3,4,5,6}

we pick 5 with a probability of P(f = 5) = 1/4, 
the probability of picking 5 as the second 
P(s = 5 | f = 5) = 0

we pick any other number except 5 and 2 as the first digit with P(f !=5) = 3/4 (this should be P(f !=5, f!=2), i'm just lazy and i'll write it like this)
the probability of picking 5 as the second is
P(s = 5 | f != 5) = 1/4 
ex. we pick 3 as the first digit --> than we have left {2,4,5,6} --> so we have 1 in 4 possible chances to get 5

now the total probability to pick 5 as the second digit is equal to the sum of the two terms:
+the probability of getting 5 as the first AND 5 as the second P(f = 5 , s = 5)
+the probability of getting anything but 5 as the first AND 5 as the second P(f != 5, s = 5)

P(f = 5 , s = 5) = P(s = 5 | f =5) * P(f = 5) = 0 * 1/4

P(f != 5 , s = 5) = P(s = 5 | f !=5) * P(f != 5) = 1/4 * 3/4

P(s = 5) = P(f = 5 , s = 5) + P(f != 5 , s = 5) = 0*1/4 + 1/4*3/4 = 3/16

3+16 = 19

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Let S be the set of permutations of the sequence 2, 3, 4, 5, 6 for whi

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