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# A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,

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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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20 Oct 2015, 03:09
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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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20 Oct 2015, 03:55
48
15
The first digit can be filled in 8 ways
For second digit , it can either be 0 or 1
Case 1 -
If second digit is 1 ,Third digit can take 10 values
number of codes = 8 * 1 * 10 = 80

Case 2 -
If second digit is 0,Third digit can take 9 values ( Third digit can't be zero)
number of codes = 8 * 1 * 9= 72

Total number of codes = 152

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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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20 Oct 2015, 04:19
10
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

Total digits = 10

Code digits = X Y Z

Now Slot X can take 8 values except 0 & 1

Slot Y can take 1 value, either 0 or 1

Then, if Slot Y takes digit "0" then slot Z can only take 9 digits (It can't take digit zero)

If slot Y takes digit "1" then slot Z can take all digits i.e, all 10 digits.

Therefore possible different codes are = 8*1*9+8*1*10 = 72+80=152 Option B

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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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10 Oct 2016, 08:10
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Hi everyone,

This is my video explanation of the question. Hope you enjoy!

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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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Updated on: 24 Sep 2016, 14:26
4
Taking two cases :

1st where the center digit is 0 ; no of possibilities = 8*1*9

2nd where the center digit is 1; no of possibilites = 8*1*10

Total = 8*1*19 = 152

Originally posted by rishi02 on 15 Jun 2016, 23:50.
Last edited by rishi02 on 24 Sep 2016, 14:26, edited 1 time in total.
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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20 Oct 2015, 06:28
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Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

Split it up in 2 scenarios:

First scenario, second digit = 1: 8*1*10 = 80 (The first digit has 8 possible numbers, the second is one and the last can be any number since the second is one)
Second scenario, second digit = 0: 8*1*9 = 72 (The first digit has again 8 possible numbers, the second is 0 in this case and therefore the last can only have 9 other digits)

Since this is an "OR" relationship, add the two 80+72 = 152

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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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24 Sep 2016, 11:16
2
Total possibilities : 8 x 2 x 10= 160
Possibilities with zero in second and third places are 8x1x1=8
Ans 160-8 = 152
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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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12 May 2018, 18:07
2
dave13 wrote:
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

YAY! my solution:)

$$C^3_{8}=56$$ first digit cant be 0, or 1
You are trying to account here for . . . three slots. See below.
$$C^1_{8}=8$$

$$C^2_{9}=36*2=72$$ multiply 36 by 2 since the second digit must be 0 or 1
$$C^1_{9}=9$$ third digits cannot both be 0 in the same code
$$56+72+36 = 164$$ i know its wrong answer...

generis, pushpitkc hello deep thinkers i tried to tackle the above question using combinatorics formula but something went wrong

non of the above mentioned solutions presented in this thread feature combinatorics formula such as the one used by me...., at least i see it that way people just use single line multiplication 8*1*10 etc ...is it another combinatorics formula that i am not aware of ? i ask this question because i dont understand based on which formula do they apply this technique ? whats logic
could you please correct my solution using combinatorics formula and suggest what is name of formula such as single line multiplication

many thanks! hope you are enjoying weekend

Hi dave13 - combinations are unwieldy here, but
they are familiar to you. Understandable.

I am pleased to acquaint you with the Fundamental Counting Principle.

(I will PM you one way I think you can use
combinations, pace my sanity)

• Fundamental Counting Principle
FCP, see below, is a LOT easier.

The "line multiplication" to which you refer
is called the Fundamental Counting Principle (FCP)

FCP underlies ALL combinatorics. The posters above use it.

FCP is often called the "slot method," or more rarely, the "line method."

You MUST know about it. Bossy bear here.

For each slot, we decide how many choices we have.
____ ____ ____

Then we multiply (if there are X ways of doing Action A,
and there are Y ways of doing Action B, then there are
X * Y ways of doing Actions A AND B)

• FCP and this problem

FIRST SLOT - How many of the 10 choices? No 0 or 1. Choices: 8
__8__ ____ ____

SECOND SLOT? This position is more restrictive than slot #1
(the outcome of third position is more specific, but not more restrictive)

That slot has 2 choices: slot MUST be 0 OR 1

How do we handle this restriction? In one of two ways.
(1) Split the cases
(2) Find all choices, subtract impermissible

TWO METHODS after the first slot

Method # 1 Split the possible cases
TWO SETS: 1 is the second digit. 0 is the second digit.

How many choices do we have for the FIRST SLOT?
10 digits total, but no 1 or 0. We have 8 choices
First slot is __8___ in BOTH cases

(1A) If "1" is in the second slot
there is ONE choice for that slot
__8__*__1__*____
Third slot? ALL 10 digits are possible
__8__ *__1__ *__10__ = 80 possible combos

(1B) If "0" is in the second slot
there is ONE choice for that slot
__8__*__1__*____

Third slot if 0 is in the second slot? How many choices?
If the second slot is 0, the third slot CANNOT be 0
We have 9 choices for 3rd slot
___8___ *___1___* ___9___ = 72 possible combos

Now what? Add or multiply or neither?

Probability rule re OR (0 OR 1, mutually exclusive) tells us

Answer choices: too small for multiplying 72 * 80

ADD. (80 + 72) = 152 possible lock combinations

Method #2: (ALL possible) - (impermissible)

(2A) ALL, adhering to the second slot rule
ignoring (no #00) rule. Possibilities for each slot?

__8__ (no 0 or 1)
__2__ (only 0 and 1)
__10_ (pretend the third slot is not restricted yet)
__8__*__2__*__10_ = 160

(2B) IMPERMISSIBLE - How many codes WILL have #00?
Think in "MUST BE" logic
How many ways can we get a #00
Slot 2: "0" = ONE choice
Slot 3: "0" = ONE choice

__8__*__1__*__1__ = 8 impermissible cases

(ALL) - (IMPERMISSIBLE): (160 - 8) = 152

Hope that helps.

In addition to resources pushpitkc listed, because I know combinatorics is not taught well and often not taught at all in certain regions,
I have added many. #1 ; #2 ; #3 Bunuel 's post pf March 29, 2017 ; #4 ; #5 (please check first two links in this post); #6 ; #7 ; #8 ; #9

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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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05 Apr 2019, 10:13
2
In the answer choises i see a lot of calculations that are IMO not needed.
The total number of possible combinations is 8*2*10=160. From those combinatios we must just remove those that have the same digit in second and third place: 8*1*1=8. Thus, 160-8=152.

Hope this helps.
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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20 Oct 2015, 22:14
1
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

we have two cases here

case 1 when 2nd digit is 0, then we can have 9 different digits in third place and 8 different digits on 1st place

no of possibilities = 9 * 8 = 72

case 2 when 2nd digit is 1, then we can have 10 different digits in third place and 8 different digits on 1st place

no of possibilities = 10 * 8 = 80

total = 72 + 80 = 152

kudos, if you like the post
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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21 May 2017, 19:59
1
Top Contributor
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

1. Number of possibilities for the first digit is 8
2. Case 1 is when the second digit is 0, number of possibilities for the third digit is 9, for a total of 8*9
3. Case 2 is when second digit is 1, number of possibilities for the third digit is 10, for a total of 8*10
4. Total number of possibilities is (2) + (3) =152
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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12 May 2018, 15:01
1
dave13 wrote:
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

YAY! my solution:)

$$C^3_{8}=56$$ first digit cant be 0, or 1

$$C^2_{9}=36*2=72$$ multiply 36 by 2 since the second digit must be 0 or 1

$$C^1_{9}=9$$ third digits cannot both be 0 in the same code

$$56+72+36 = 164$$ i know its wrong answer...

generis, pushpitkc hello deep thinkers i tried to tackle the above question using combinatorics formula but something went wrong

non of the above mentioned solutions presented in this thread feature combinatorics formula such as the one used by me...., at least i see it that way people just use single line multiplication 8*1*10 etc ...is it another combinatorics formula that i am not aware of ? i ask this question because i dont understand based on which formula do they apply this technique ? whats logic
could you please correct my solution using combinatorics formula and suggest what is name of formula such as single line multiplication

many thanks!
hope you are enjoying weekend

Hi dave13

The combinatorics formula wouldn't work in problems like these. Also, when you wrote
your solution why did you use $$C^3_{8}$$ for the first digit and $$C^2_{9}$$ for the second digit when
you only had to choose one digit?

You could look at the solution given by Skywalker18 to understand how problems like
these should be solved. Here is a problem from the GMATClub Quantitative Mega-thread
which discusses why we can't use combinatorics formula for all questions
https://gmatclub.com/forum/combinatoric ... l#p1579515

Hope this helps you.
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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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08 Nov 2015, 08:17
Result is difference of all possible choices, minus the choices where second digit is 0 and third digit is 0.

1)All choices

On the first place you can have 8 different digits. on second you can have 2 digits and on the third you can have 10 digits.
So, 8*2*10=160

2)Second and third places are 0

On the first place you can still have 8 different digits, on the second you can have only one possible coice (0) and on the third you again have only one possible choice (0).
So, 8*1*1=8

Difference is 160-8=152

regards
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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22 May 2017, 12:07
First Digit can be: 2,3,4,5,6,7,8,9
Second Digit can be: 0,1
Third Digit can be: 0,1,2,3,4,5,6,7,8,9

Case 1: If Second Digit is 0

8 * 1 * 9 = 72

Case 2: If Second Digit is 1

8 * 1 * 10 = 80

Total = 72+80 = 152

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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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12 Sep 2017, 02:48
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

8*1*9+8*1*10 = 72+80=152
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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12 Sep 2017, 03:24
First place can take 8 digits, second digit can take 2 digits(0,1) and third place can take any of the digits.
2 cases possible :
Case 1 : When zero is present in second place - 8*1*9 = 72
Case 2 : When 1 is present in second place - 8*1*10 = 80

Total = 152

Option B
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A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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12 May 2018, 04:39
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

YAY! my solution:)

$$C^3_{8}=56$$ first digit cant be 0, or 1

$$C^2_{9}=36*2=72$$ multiply 36 by 2 since the second digit must be 0 or 1

$$C^1_{9}=9$$ third digits cannot both be 0 in the same code

$$56+72+36 = 164$$ i know its wrong answer...

generis, pushpitkc hello deep thinkers i tried to tackle the above question using combinatorics formula but something went wrong

non of the above mentioned solutions presented in this thread feature combinatorics formula such as the one used by me...., at least i see it that way people just use single line multiplication 8*1*10 etc ...is it another combinatorics formula that i am not aware of ? i ask this question because i dont understand based on which formula do they apply this technique ? what`s logic
could you please correct my solution using combinatorics formula and suggest what is name of formula such as single line multiplication

many thanks!
hope you are enjoying weekend
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Joined: 03 Aug 2017
Posts: 102
Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,  [#permalink]

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27 Nov 2019, 04:33
Bunuel wrote:
A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 according to the following constraints. The first digit cannot be 0 or 1, the second digit must be 0 or 1, and the second and third digits cannot both be 0 in the same code. How many different codes are possible?

(A) 144
(B) 152
(C) 160
(D) 168
(E) 176

Kudos for a correct solution.

we have two cases here

case 1 when 2nd digit is 0, then we can have 9 different digits in third place and 8 different digits on 1st place

no of possibilities = 8*1*9 = 72 ( The 1st spot can be taken by all nos except 0, and 1 ( SO 8 ways to do it ) 2nd spot is taken by 0 ( 1 way to do it ) and third spot can be taken by all other nos expect 0 ie 9 ways )

case 2 when 2nd digit is 1, then we can have 10 different digits in third place and 8 different digits on 1st place
( The 1st spot can be taken by all nos except 0, and 1 ( SO 8 ways to do it ) 2nd spot is taken by 1 ( 1 way to do it ) and the third spot can be taken by any no ie 10 ways )

no of possibilities = 8* 1* 10 = 80 ways

since its an OR situation we need to add both scenarios

total = 72 + 80 = 152

Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5,   [#permalink] 27 Nov 2019, 04:33
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