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.

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Hope it helps.
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If n is the number of distinct letters used to create the two lettered codes, then a total of \(n * n = n^2\) different codes can be created. We need \(n^2\geq60\). The smallest n which fulfills this condition is n = 8.

Answer C
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Hey Bunuel,

One doubt regarding this question. Doesn't the phrase "ordered pair" mean nothing in the question? My inference was as it is an ordered pair it should be in alphabetical order.

Please clarify. Thanks.

An ordered pair of letters mean that code AB considered different from code BA, so both are possible.
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I came up with 2P# + 1C# >= 60
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If we assume code length to be 3 i.e. ABC,AAA,AAB...SO on.. minimum number of letters required would be 4.
n^3 >=60 =>n =4

Question is framed incorrectly because it has not mentioned the code length.
If length is 2, 8 will be the answer.If length is 3 ,4 will be the answer.

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?

Pair means two.
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Hi Bunuel,
If the prompt restricts the duplication of a letter, can I use permutation in that case? if yes, we need more than 8 so 8p2=72.

Is it correct?

Thanks

Hi ,
if the prompt restricts the usage
first place can be filled by n letters and 2nd place by n-1 letters..
so total ways n(n-1), which same as np2..
so you are correct....
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Pls correct me,
I got 6 no of letters for 60 maximum no. of Items when the no. of letters is not mentioned for the code,


since in combination ORDER does not matter and when we place SOME digits in a different order in Permutations, ONLY one out of them is in ascending order, we can work on Combinations
say total n digits are required
single digit will be nC1...
2 digits - nC2
3 digits= nC3...and so on..
so we are looking for nC1+nC2+nC3+...nCn≥60
nC1+nC2+nC3+...nCn≥60
now,
nC0+nC1+nC2+nC3+...nCn=2^n
nC0+nC1+nC2+nC3+...nCn=2^n is a formula..

so nC1+nC2+nC3+...nCn=2^n−nC0=2^n−1
nC1+nC2+nC3+...nCn=2n−nC0=2^n−1..
so 2^n−1≥60.................2^n≥61...................so..n≥6

so n=6 will be the minimum no. of codes that can be used for arbitrary no of position in the sequential order.

pls correct if if anything wrong in the logic

We can let n = the number of letters needed to make the codes. Since we can use the same letter for the second letter and the first letter, we have n choices for the first letter and n choices for the second letter also. Thus, the number of codes we can make is n x n = n^2, and we want this to be greater than or equal to 60. That is, n^2 ≥ 60.

We see that the smallest integer value of n must be 8 in order for n^2 ≥ 60; thus, the minimum number of letters that must be used is 8.

Answer: C
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Hello Bunuel
I thought particular order means a single order in which AB and BA both are not acceptable. So, I assumed the answer would be nC2≥60.

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.

The key to solving the above equation is getting the combination right
I guess the order is alphabhetical therefore
let us take the possiblity of 8(even though it's brute force we will get the idea of the same and afterwards we can decide )
AB,A_=>since there are seven characters that 7(6)/2= 21
B_=> since there are 6 eleminating A we have 6*5/2= 15
C_=> since there are 5 eleminating A,B we have 5*4/2=10
D_=>since there are 4 eleminating A,B,C we have 4*3/2=6
E_=>since there are 4 eleminating A,B,C ,D we have 3
Total 55
other possiblities are
FG,FH,GH, GI, HI =60
Exactly what we were looking along with alphabhetical a very tight question brute force helped may not be so useful in GMAT
Hence IMO C

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