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

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Joined: 01 Sep 2010
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How many 4 digit codes can be made, if each code can only [#permalink]

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04 Oct 2010, 06:00
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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).
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 07:44, edited 1 time in total.
RENAMED THE TOPIC.
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04 Oct 2010, 07:41
1
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Expert's post
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).

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: $$4^4+4^2+3*4^3=464$$.

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04 Oct 2010, 07:45
2
KUDOS
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).

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

PS : Some others also deserve thanks
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04 Oct 2010, 08:41
Hi guys,
Thanks you BOTH for the explanation. It is clear now.
You got +1 from me.
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Joined: 26 Mar 2010
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05 Oct 2010, 12:33
Bunuel 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).

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: $$4^4+4^2+3*4^3=464$$.

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???
Math Expert
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Posts: 37148
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Kudos [?]: 96818 [0], given: 10786

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05 Oct 2010, 12:37
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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05 Oct 2010, 12:53
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...
Math Expert
Joined: 02 Sep 2009
Posts: 37148
Followers: 7274

Kudos [?]: 96818 [1] , given: 10786

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05 Oct 2010, 13:01
1
KUDOS
Expert's post
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.
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05 Oct 2010, 13:59
Thanks Bunuel... +1 ... u already have many I knw
Re: 4 digit codes   [#permalink] 05 Oct 2010, 13:59
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