GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Dec 2019, 06:52 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59728
Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  [#permalink]

### Show Tags 00:00

Difficulty:   55% (hard)

Question Stats: 43% (02:15) correct 57% (01:25) wrong based on 14 sessions

### HideShow timer Statistics

Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for which the first term is not 1. A permutation is chosen randomly from S. The probability that the second term is 2, in lowest terms, is a/b. What is a + b?

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

_________________
e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  [#permalink]

### Show Tags

Solution

Given:
• S denotes the set of permutations of the sequence 1, 2, 3, 4, 5 for which the first term is not 1.
• A permutation is chosen randomly from S.
• The probability that the second term is 2, in lowest terms, is a/b.

To find:
• The value of a + b.

Approach and Working:
As the first term is not 1,
• Total number of ways the first term can be chosen = 4
• And, the number of possible permutations for each of the first terms = 4!
• Hence, total permutations = 4 x 4!

If the second term is 2,
• Total number of ways the first term can be chosen = 3
• And, the number of possible permutations = 3!
• Hence, total permutations = 3 x 3!

So, the value of a/b = (3 * 3!)/(4 * 4!) = 18/96 = 3/16
• The value of a + b = 3 + 16 = 19

Hence, the correct answer is option E.

_________________
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 5484
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  [#permalink]

### Show Tags

EgmatQuantExpert
Could you please explain on why you have done
So, the value of a/b = (3 * 3!)/(4 * 4!) = 18/96 = 3/16
• The value of a + b = 3 + 16 = 19

EgmatQuantExpert wrote:

Solution

Given:
• S denotes the set of permutations of the sequence 1, 2, 3, 4, 5 for which the first term is not 1.
• A permutation is chosen randomly from S.
• The probability that the second term is 2, in lowest terms, is a/b.

To find:
• The value of a + b.

Approach and Working:
As the first term is not 1,
• Total number of ways the first term can be chosen = 4
• And, the number of possible permutations for each of the first terms = 4!
• Hence, total permutations = 4 x 4!

If the second term is 2,
• Total number of ways the first term can be chosen = 3
• And, the number of possible permutations = 3!
• Hence, total permutations = 3 x 3!

So, the value of a/b = (3 * 3!)/(4 * 4!) = 18/96 = 3/16
• The value of a + b = 3 + 16 = 19

Hence, the correct answer is option E.

e-GMAT Representative V
Joined: 04 Jan 2015
Posts: 3158
Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  [#permalink]

### Show Tags

Archit3110 wrote:
EgmatQuantExpert
Could you please explain on why you have done
So, the value of a/b = (3 * 3!)/(4 * 4!) = 18/96 = 3/16
• The value of a + b = 3 + 16 = 19

Hey Archit3110,
in the question, a/b is defined as the probability of choosing a set with second term 2.

Now, number of ways we can select a set with second term 2 = 3 * 3!
And, total number of ways we can select a set = 4 * 4!
Hence, the probability = $$\frac{a}{b} = \frac{3 * 3!}{4 * 4!}$$

In lowest term, the value of the fraction becomes $$\frac{3}{16}$$
Hence, a + b = 3 + 16 = 19

_________________ Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi   [#permalink] 21 Mar 2019, 01:46
Display posts from previous: Sort by

# Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  