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How many 4 digit codes can be made, if each code can only [#permalink]
04 Oct 2010, 06:00

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Difficulty:

5% (low)

Question Stats:

50% (02:40) correct
50% (01:30) wrong based on 6 sessions

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.

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;

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

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???

(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.

(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...

(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}