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

Question

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.

Bunuel · Answer

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.

shrouded1 · Answer

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:

utin · Answer

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???

Bunuel · Answer

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.

utin · Answer

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...

Bunuel · Answer

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.

utin · Answer

Thanks Bunuel... +1 ... u already have many I knw

shashankism · Answer

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

testcracker · Answer

great solution as always, man

fugitsapiente · Answer

question: among the 464 possible codes; one of the formed codes is 1933 which is a 4 digit prime bigger than 20. this goes against what the question asks; right?

we can only use prime numbers less than 20; right?

another example- any code that contains digits 2 and 3 together should be disqualified since 23 is a prime that is greater than 20.

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

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