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

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

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New post 20 Mar 2019, 23:28
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

27% (03:01) correct 73% (01:25) wrong based on 11 sessions

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Joined: 04 Jan 2015
Posts: 2808
Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi  [#permalink]

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New post 21 Mar 2019, 00:04

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.

Answer: E

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

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New post 21 Mar 2019, 01:35
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.

Answer: E

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

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New post 21 Mar 2019, 01:46
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

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Re: Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi   [#permalink] 21 Mar 2019, 01:46
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Let S be the set of permutations of the sequence 1, 2, 3, 4, 5 for whi

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