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

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

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

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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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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]
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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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, [#permalink]
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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]
Total numbers: 0 to 9 = 10.

Three-digit code: _, _, _

The first digit cannot be 0 or 1: First space will have 8 choices

The second digit must be 0 or 1:

Case I: If we fix '0', then the third space cannot be '0', and hence third space will have only 9 choices.

=> Total ways: 8 * 1 * 9 = 72

Case II: If we fix '1', then the third space can have any value, and hence third space will have 10 choices.

=> Total ways: 8 * 1 * 10 = 80

Overall ways: 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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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

Solution:

We see that the first digit has 8 choices, the second has 2, and the third has 10. Therefore, there are 8 x 2 x 10 = 160 possible 3-digit codes. However, this includes codes such as 200, 300, …, and 900, which are not allowed. Therefore, there are actually 160 - 8 = 152 possible codes.

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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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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

Video solution by GMATinsight

­
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, [#permalink]
KarishmaB Bunuel
I need some clarity on the approach
All - favourable cases approach = 152

To calculate All cases, we consider 8 x 2 x 10 = 160
Why don't we remove all restriction on each digit for all cases (first position will have 10 option, similarly second position will have 10 option as we can repeat the digit). In other words, why are we restricting 1st & 2nd digit and not 3rd digit for all cases?
What's wrong with this approach?
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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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Sneha2021 wrote:
KarishmaB Bunuel
I need some clarity on the approach
All - favourable cases approach = 152

To calculate All cases, we consider 8 x 2 x 10 = 160
Why don't we remove all restriction on each digit for all cases (first position will have 10 option, similarly second position will have 10 option as we can repeat the digit). In other words, why are we restricting 1st & 2nd digit and not 3rd digit for all cases?
What's wrong with this approach?

All 10 digits are not allowed in first position. Only 8 are allowed. Similarly only 2 digits are allowed in second place. We can easily adjust our total cases for these constraints by using 8*2*10. Then the only thing not allowed is the second-third digit combination of 00 and we have 8 such cases so we remove them.
We cannot adjust for this directly (because third digit that is not allowed depends on what second digit is) and hence we do Total - the ones not allowed.

If we start with 10*10*10, we need to remove too many constraints and that's unnecessary work. First digit, 0 and 1 not allowed so remove 2*10*10. We are left with 800. Second digit only 0 and 1 are allowed so remove 8*8*10 so we are left with 160.
Now remove 0-0 combination to be left with 152. Too much work.

Alternatively, we can just add the cases that are allowed i.e. 8*1*10 + 8*1*9 = 152
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Re: A three-digit code for certain locks uses the digits 0, 1, 2, 3, 4, 5, [#permalink]
ScottTargetTestPrep 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

Solution:

We see that the first digit has 8 choices, the second has 2, and the third has 10. Therefore, there are 8 x 2 x 10 = 160 possible 3-digit codes. However, this includes codes such as 200, 300, …, and 900, which are not allowed. Therefore, there are actually 160 - 8 = 152 possible codes.

ScottTargetTestPrep
To confirm, you do not need to subject out 100 because multiplying by 8 to start off with already disregards 100 as a possibility? Thank you!
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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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woohoo921 wrote:
ScottTargetTestPrep 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

Solution:

We see that the first digit has 8 choices, the second has 2, and the third has 10. Therefore, there are 8 x 2 x 10 = 160 possible 3-digit codes. However, this includes codes such as 200, 300, …, and 900, which are not allowed. Therefore, there are actually 160 - 8 = 152 possible codes.