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An unfortunate thief was caught while trying to decode the 3-digit num

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An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

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New post 29 May 2018, 01:51
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

73% (01:11) correct 27% (01:35) wrong based on 93 sessions

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An unfortunate thief was caught while trying to decode the 3-digit numerical code of a lock. The Police caught him while he was making the second-last possible unique attempt. How many attempts did the thief make in total?

A. 5
B. 26
C. 719
D. 999
E. 1000


To solve question 4: P&C Practice Question 4

Related Articles:

Article-1: Learn when to “Add” and “Multiply” in Permutation & Combination questions
Article-2: Fool-proof method to Differentiate between Permutation & Combination Questions
Article-3: 3 deadly mistakes you must avoid in Permutation & Combination

All Articles: Must Read Articles and Practice Questions to score Q51 !!!!

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An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

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New post 29 May 2018, 05:40
EgmatQuantExpert wrote:
An unfortunate thief was caught while trying to decode the 3-digit numerical code of a lock. The Police caught him while he was making the second-last possible unique attempt. How many attempts did the thief make in total?

A. 5
B. 26
C. 719
D. 999
E. 1000


We are talking about the 3-digit numerical code. There are 10^3 or 1000 possibilities.
Since the thief is making his second-last possible unique attempt, he would be making
his 1000-1 = 999 attempt.

Therefore, the thief will be making his 999th attempt(Option D)
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Re: An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

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New post 31 May 2018, 01:53

Solution



Given:
    • The thief was trying to decode a 3-digit numerical code of a lock
    • He was caught while making the second-last possible event

To find:
    • In total, how many attempts did the thief make

Approach and Working:
    • As it is a 3-digit numerical code, every place in the code can be filled by 10 possible digits
    • Therefore, total number of possible codes = 10 * 10 * 10 = 1000
However, among all the 1000 codes, only 1 is correct
    • Therefore, the maximum number of unsuccessful attempts = second-last possible unique attempts = 1000 – 1 = 999

Hence, the correct answer is option D.

Answer: D
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Re: An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

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New post 09 Aug 2019, 22:53
EgmatQuantExpert wrote:

Solution



Given:
    • The thief was trying to decode a 3-digit numerical code of a lock
    • He was caught while making the second-last possible event

To find:
    • In total, how many attempts did the thief make

Approach and Working:
    • As it is a 3-digit numerical code, every place in the code can be filled by 10 possible digits
    • Therefore, total number of possible codes = 10 * 10 * 10 = 1000
However, among all the 1000 codes, only 1 is correct
    • Therefore, the maximum number of unsuccessful attempts = second-last possible unique attempts = 1000 – 1 = 999

Hence, the correct answer is option D.

Answer: D


Hi,

I have a questions regarding this. When we say 10x10x10 will this method count e.g a combination of 322 as one of two possible identical choices ? I mean will it consider the two 2s as different or as same ? My guess is that since this is a permutation based approach it will count the two 322s as different. Similarly the result should contain three combinations of 333s. Is this true ?

Will appreciate a response on this.
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Re: An unfortunate thief was caught while trying to decode the 3-digit num   [#permalink] 09 Aug 2019, 22:53
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An unfortunate thief was caught while trying to decode the 3-digit num

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