Bunuel wrote:
thebogie17 wrote:
In a certain appliance store, each model of television is uniquely designated by a code made up of a particular ordered pair of letters. If the store has 60 different models of televisions, what is the minimum number of letters that must be used to make the codes?
A. 6
B. 7
C. 8
D. 9
E. 10
I think the official answer supplied by PR is incorrect.
I Feel that D is the correct answer, they say C.
Am I missing something?
Notice that we are not told that letters in two-letter code must be different. For example three letters A, B, and C give the following codes:
AA;
BB;
CC;
AB;
BA;
AC;
CA;
BC;
CB.
So, if we have \(n\) distinct letters, then we can make \(n^2\) different codes (since each X in XX code can take \(n\) values). As there are 60 different models of TV then \(n^2\geq{60}\) must hold true. Since \(n\) must be an integer then the least value of \(n\) is 8.
Answer: C.
Hope it helps.
Hi Bunuel,
Can you pls help me to understand, why are we not supposed to calculate for doubles here?
For example AB and BA are different, that's ok. We count them as 2.
But, AA and AA (or like this A1A2 and A2A1) are the same, but we are not counting for doubles and not dividing by 2!
Why is it the case here, but not in the others?
For example, there was a question with word ILLUSION, where we should calculate possibilities of different combinations from this letters and we calculated it like 8!/2!2! (because there two I's and L's).
Thank you in advance.