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

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

Answer B
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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
that we add the possibilities

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

Answer B

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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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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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

Answer B

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

Answer choice B

kudos, if you like the post

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
Answer B.

regards

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

Total possibilities : 8 x 2 x 10= 160
Possibilities with zero in second and third places are 8x1x1=8
Ans 160-8 = 152

Hi everyone,

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



Rowan

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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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

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.

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

Answer B

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.

Answer: B
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Answer: Option B

Video solution by GMATinsight



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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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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!

Yes. Exactly.
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