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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