GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 31 May 2020, 02:32

### 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

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

Author Message
TAGS:

### Hide Tags

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3386
An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

### Show Tags

29 May 2018, 00:51
00:00

Difficulty:

25% (medium)

Question Stats:

70% (01:12) correct 30% (01:37) wrong based on 121 sessions

### HideShow timer Statistics

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 !!!!

_________________
e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3386
Re: An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

### Show Tags

31 May 2018, 00:53
1

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.

_________________
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3253
Location: India
GPA: 3.12
An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

### Show Tags

29 May 2018, 04:40
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)
_________________
You've got what it takes, but it will take everything you've got
Manager
Joined: 27 Mar 2017
Posts: 153
Re: An unfortunate thief was caught while trying to decode the 3-digit num  [#permalink]

### Show Tags

09 Aug 2019, 21:53
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.

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.
Re: An unfortunate thief was caught while trying to decode the 3-digit num   [#permalink] 09 Aug 2019, 21:53