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How many 4 digit codes can be made, if each code can only contain

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How many 4 digit codes can be made, if each code can only contain [#permalink]

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How many 4 digit codes can be made, if each code can only contain prime numbers that are less than 20?

A. 24
B. 102
C. 464
D. 656
E. 5040

Hi all,

As you saw, I have published a bunch of questions in the past hour. Most of these questions are taken from a collection of hard quantitative questions provided by The Princeton Review (a.k.a - Killer Math).

I have posted any question that I have solved incorrectly, either due to careless error or concept error, in order to share them with everyone here.

Thank you all for the help and detailed explanations (especially you - Bunuel).

It is very helpful.
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 May 2017, 08:48, edited 2 times in total.
RENAMED THE TOPIC.

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 04 Oct 2010, 08:41
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eladshush wrote:
Hi all,

As you saw, I have published a bunch of questions in the past hour. Most of these questions are taken from a collection of hard quantitative questions provided by The Princeton Review (a.k.a - Killer Math).

I have posted any question that I have solved incorrectly, either due to careless error or concept error, in order to share them with everyone here.

Please consider the following problem that I am not sure I understood:

How many 4 digit codes can be made, if each code can only contain prime numbers that are less than 20?

A. 24
B. 102
C. 464
D. 656
E. 5040

Thank you all for the help and detailed explanations (especially you - Bunuel).
It is very helpful.


The question is a little bit ambiguous but I think it means the following:

I guess as it's not mentioned primes can be repeated.

There are:
4 one digit primes (O) less than 20 - 2, 3, 5, 7;
4 two digit primes (T) less than 20 - 11, 13, 17, 19;

Thus 4-digit number could be of the following type:

OOOO, for example: 2|3|5|7 or 2|2|7|7. Each O can take 4 values from {2, 3, 5, 7}, so total ways for this type is \(4^4\);

TT, for example: 11|11 or 19|17. Each T can take 4 values from {11, 13, 17, 19}, so total ways for this type is \(4^2\);

TOO, for example: 11|3|5 or 19|7|2. T can take 4 values from {11, 13, 17, 19} and each O can take 4 values from {2, 3, 5, 7}, so total ways for this type is \(4*4^2=4^3\);
OTO, for example: 2|13|5 or 7|19|2. The same as above: \(4^3\);
OOT, for example: 2|5|19 or 7|2|17. The same as above: \(4^3\);

Total: \(4^4+4^2+3*4^3=464\).

Answer: C.
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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 04 Oct 2010, 08:45
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eladshush wrote:
Hi all,

As you saw, I have published a bunch of questions in the past hour. Most of these questions are taken from a collection of hard quantitative questions provided by The Princeton Review (a.k.a - Killer Math).

I have posted any question that I have solved incorrectly, either due to careless error or concept error, in order to share them with everyone here.

Please consider the following problem that I am not sure I understood:

How many 4 digit codes can be made, if each code can only contain prime numbers that are less than 20?

A. 24
B. 102
C. 464
D. 656
E. 5040

Thank you all for the help and detailed explanations (especially you - Bunuel).
It is very helpful.


First note all the single digit primes {2,3,5,7}
And then all the 2-digit ones {11,13,17,19}

Case 1
Codes formed with 2 two digit primes
(2-digit prime) (2-digit prime)
No of ways = 4x4 = 16

Case 2
Codes formed with 4 one digit primes
(1-digit prime) (1-digit prime) (1-digit prime) (1-digit prime)
No of ways = 4x4x4x4 = 256

Case 3
Codes formed with 2 one-digit primes and 1 two-digit prime
(1-digit prime) (1-digit prime) (2-digit prime)
(1-digit prime) (2-digit prime) (1-digit prime)
(2-digit prime) (1-digit prime) (1-digit prime)
Each set can be formed in 4x4x4 ways
So total = 3x64 = 192


Total number = 192+256+16 = 464

Answer is (c)

PS : Some others also deserve thanks :wink:
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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 05 Oct 2010, 13:33
Bunuel wrote:
eladshush wrote:
Hi all,

As you saw, I have published a bunch of questions in the past hour. Most of these questions are taken from a collection of hard quantitative questions provided by The Princeton Review (a.k.a - Killer Math).

I have posted any question that I have solved incorrectly, either due to careless error or concept error, in order to share them with everyone here.

Please consider the following problem that I am not sure I understood:

How many 4 digit codes can be made, if each code can only contain prime numbers that are less than 20?

A. 24
B. 102
C. 464
D. 656
E. 5040

Thank you all for the help and detailed explanations (especially you - Bunuel).
It is very helpful.


The question is a little bit ambiguous but I think it means the following:

I guess as it's not mentioned primes can be repeated.

There are:
4 one digit primes (O) less than 20 - 2, 3, 5, 7;
4 two digit primes (T) less than 20 - 11, 13, 17, 19;

Thus 4-digit number could be of the following type:

OOOO, for example: 2|3|5|7 or 2|2|7|7. Each O can take 4 values from {2, 3, 5, 7}, so total ways for this type is \(4^4\);

TT, for example: 11|11 or 19|17. Each T can take 4 values from {11, 13, 17, 19}, so total ways for this type is \(4^2\);

TOO, for example: 11|3|5 or 19|7|2. T can take 4 values from {11, 13, 17, 19} and each O can take 4 values from {2, 3, 5, 7}, so total ways for this type is \(4*4^2=4^3\);
OTO, for example: 2|13|5 or 7|19|2. The same as above: \(4^3\);
OOT, for example: 2|5|19 or 7|2|17. The same as above: \(4^3\);

Total: \(4^4+4^2+3*4^3=464\).

Answer: C.



Hi Bunuel,

why can't i write TOO,OTO,OOT AS

(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 05 Oct 2010, 13:37
utin wrote:
Hi Bunuel,

why can't i write TOO,OTO,OOT AS

(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???



Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be \(4^3*\frac{3!}{2!}=4^3*3\).

Hope it's clear.
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 05 Oct 2010, 13:53
Bunuel wrote:
utin wrote:
Hi Bunuel,

why can't i write TOO,OTO,OOT AS

(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???



Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be \(4^3*\frac{3!}{2!}=4^3*3\).

Hope it's clear.



I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!


this might clear my entire probability confusion i hope... :)

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 05 Oct 2010, 14:01
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utin wrote:
Bunuel wrote:
utin wrote:
Hi Bunuel,

why can't i write TOO,OTO,OOT AS

(4^3)*3! , taking the T as one entity ans assuming that 3 things can be arranged in 3! ways???



Just one thing: TOO can be arranged in 3!/2! ways and not in 3! (# of permutations of 3 letters out which 2 O's are identical is 3!/2!), so it would be \(4^3*\frac{3!}{2!}=4^3*3\).

Hope it's clear.



I though about the same but but when i see that TOO as three things to be arranged in 3! ways then i also thought that OO ARE TWO DIGITS AND THEY ARE TWO DIFFERENT PRIME NOS SO WHY DIVIDE BY 2!


this might clear my entire probability confusion i hope... :)


First of all two 1-digit primes can be the same, but it's not important here.

We are counting # of ways 4-digit number can be formed with two 1-digit primes and one 2-digit prime:
{1-digit}{1-digit}{2-digit}
{1-digit}{2-digit}{1-digit}
{2-digit}{1-digit}{1-digit}

Total 3 ways.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 05 Oct 2010, 14:59
Thanks Bunuel... +1 ... u already have many I knw :)

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Re: How many 4 digit codes can be made, if each code can only contain [#permalink]

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New post 03 Oct 2017, 00:09
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eladshush wrote:
How many 4 digit codes can be made, if each code can only contain prime numbers that are less than 20?

A. 24
B. 102
C. 464
D. 656
E. 5040

Hi all,

As you saw, I have published a bunch of questions in the past hour. Most of these questions are taken from a collection of hard quantitative questions provided by The Princeton Review (a.k.a - Killer Math).

I have posted any question that I have solved incorrectly, either due to careless error or concept error, in order to share them with everyone here.

Thank you all for the help and detailed explanations (especially you - Bunuel).

It is very helpful.


Though the question is little bit confusing, but conceptually this question is a real gem. More details should have been provided about the restrictions on selection of digits.

So, there are 4 one digit prime numbers to be used they are {2,3,5,7}. Lets call the single digits as O
& there are 4 2-digit prime numbers to be used they are {11, 13, 15, 17}. And double digits as T.

Now there are 3 ways of selecting digits.

Case 1: All one digit prime numbers OOOO
No. of codes = 4^4 (since repetition of digit is allowed = 256

Case 2: 2 Two digit prime numbers TT
No. of codes = 4*4 = 16

Case 3 : one two digit prime number & 2 one digit prime number : OOT, OTO, TOO
No. of codes = 4*4*4 + 4*4*4 + 4*4*4 = 3*4^3 = 192

Total no. of codes = 256 + 16 + 192 = 464

Answer C
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Re: How many 4 digit codes can be made, if each code can only contain   [#permalink] 03 Oct 2017, 00:09
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