By trial and error:
If n=4, then n(n+1)=20<24;
If n=5, then n(n+1)=30>24.

Hence, \(n_{min}=5\).

Try similar questions to practice: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1296049

Hope this helps.
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I am still having problems with this question. Why do we devide the combinations formula into n(n-1)/2?
Shouldnt it be 2!/n!(2-n)! ?

\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.
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Probability and Combinatorics are my weakest subjects by far, so please ignore the rudimentary question.

When we say \(C^2_n+n\geq{12}\) that means that we are going to find a combination of 2 letters out of a group of n letters which in turn would yield "x" amount of options. Correct? If so, why are we adding the n following that equation and more importantly, how does that equation yield 5? When I factor it out, i get n(n+1) >= 24. That yields -1 and 0. Why am I so off here?

My question would be - what does this formula mean and how do you solve it? \(C^2_n+n\geq{12}\)

Also, the question is saying that they need to be in alphabetical order, doesn't that mean that order DOES matter? How does that affect the above equation.

P.S: For what it's worth, I've read the Combinatorics and Probability strategy guide(Manhattan Gmat) and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful?

EDIT: Simplifying my question.

The first advice would be, and I cannot stress this enough, to read the whole thread and follow the links to similar problems.

As for your questions:

Why we are adding n.

The question says that the code can consists of 1 or 2 letters. Now, if we have n letters how many codes we can make?

The # of single letter codes possible would be n itself;
The # of pair of distinct letters codes possible would be \(C^2_n\).

So, out of n letters we can make \(n+C^2_n\) codes: n one-letter codes and \(C^2_n\) two-letter codes.

How the equation yields 5

By trial and error:
If n=4, then n(n+1)=20<24;
If n=5, then n(n+1)=30>24.

Hence, \(n_{min}=5\).

Notice that we have \(n(n+1)\geq{24}\) NOT \(n(n+1)\geq{0}\).

About the alphabetical order.

Check here: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1150091 and here: a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html#p1296053

Similar questions to practice:
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Hope this helps.
_________________

I don't get how this is B instead of D. Aside from the formula, it says in alphabetical order, so how can you count AC, etc.?

AC is in alphabetical order while CA is not.
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Hi Bunuel,

I'm confused on when you show that n! = (n-2)!*(n-1)*n...why is n! only limited to those 3 factors? I guess the question is why do you start at (n-2)!?

Thanks.!


\(C^2_n=\frac{n!}{2!(n-2)!}\). Now, notice that \(n!=(n-2)!*(n-1)*n\), hence \(C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}\).

Hope it's clear.

n! is the product of positive integers from 1 to n, inclusive: n! = 1*2*...*(n-4)*(n-3)(n-2)(n-1)n. To simplify \(\frac{n!}{2!(n-2)!}\) I wrote n! as (n-2)!*(n-1)*n this enables us to reduce by (n-2)! to get \(\frac{(n-1)n}{2}\).

Hope it's clear.
_________________

Thanks Bunuel. I'm sorry, but I'm still not following.

Why couldn't you simplify n! as (n-4)!*(n-3)!*(n-2)!*(n-1)!*n? What tells you that n! starts with (n-2)! and not a smaller number like (n-3)! or further?

What is the significance about starting with "2" using (n-2)! and working your way down to "n" when simplifying this?

Thanks.

I think you need to brush up fundamentals.

The factorial of a non-negative integer n, denoted by \(n!\), is the product of all positive integers less than or equal to n.

For example: \(4!=1*2*3*4=24\).

Now, consider this: \(10! = 1*2*3*...*8*9*10\) can also be written as \(9!*10 = (1*2*3*...*8*9)*10 = 10!\) or \(8!*9*10 = (1*2*3*...*8)*9*10 = 10!\), or \(7!*8*9*10 = (1*2*3*...*7)*8*9*10 = 10!\) ...

Similarly \(n!\) (n factorial) can be written as \((n-2)!*(n-1)*n\) (n-2 factorial multiplied by (n-1)*n). We need to write it that way in order to be able to reduce by \((n-2)!\).

Hope it's clear.
_________________

Hi All,

In these sorts of questions, when the answer choices are relatively small, it's often fairly easy to "brute force" the correct answer and avoid complicated calculations entirely.

BrainLab's idea to just "map out" the possibilities is a relatively simple, effective approach. Since we're asked for the LEAST number of letters that will give us 12 unique codes, we start with Answer A.

If we had 4 letters: A, B, C, D

1-letter codes: A, B, C, D
2-letter alphabetical codes: AB, AC, AD, BC, BD, CD
Total Codes = 4 + 6 = 10

This result is TOO LOW.

From here, you know that we just need 2 more codes, so adding 1 more letter would give us those extra codes (and more)...but here's the proof that it happens....

If we had 5 letters: A, B, C, D, E

1-letter codes: A, B, C, D, E
2-letter alphabetical codes: AB, AC, AD, AE, BC, BD, BE, CD, CE, DE
Total Codes = 5 + 10 = 15 codes

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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Hi bunuel!

Sorry for opening the old topic but my question is why are you and everyone else writing the "c(n, k)" upside down. Have a look at the screenshot of the gmat club math book and the equation everyone is referring to.

Thanks!

\(C^1_3\), \(C^3_1\), 3C1, mean the same thing choosing 1 out of 3, just different ways to write. Can it be choosing 3 out of 1???
_________________

Hi Folks,

Pair would mean 2 elements.So we will consider single and pairs(with 2 elements) only right?

Hi kirtivardhan,

You are right, pair means 2 elements. Hence the solution considers both single letters and pair of letters for formation of the code. Let me know at which point you are having a doubt.

Regards
Harsh
_________________

Hi Harsh,
If we try to solve it by trial and error then for 5 letters it would be 5 single letters and AB,BC,CD,DE in alphabetical order.that makes the total as 9.Then how is the answer 5?

Hi kirtivardhan,

Writing the letters in alphabetical order does not mean using only consecutive alphabets. For a two alphabet combination where the first alphabet is A, every other alphabet coming after A will have the code in in alphabetical order. So, AB, AC, AD and AE are in alphabetical order. Similarly for the first alphabet as B we will have BC, BD & BE in alphabetical order. Again, for C we would have CD & CE and for D we would have DE in alphabetical order.

You missed out on the combination of these alphabets.

Another way to understand this is for the alphabet A, there are 4 other letters (i.e. B, C, D, E) which can follow A, so A will have 4 possible combinations in alphabetical order. For alphabet B, there are 3 letters (i.e. C, D, E) which can follow B, so B will have 3 possible combinations in alphabetical order. Similarly for C, we will have 2 possible combinations (i.e. D, E) and 1 possible combination (i.e. E) for D.

Total possible two alphabet combinations = 4 + 3 + 2 + 1 = 10
Total possible single alphabet combination = 5
Hence, total possible combinations i.e. distinct codes possible = 10 + 5 = 15 > 12.

Try out this exercise for 4 letters and you will observe that the total distinct possible codes is less than 12.

Hope it's clear

Regards
Harsh
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i understand the concept of the combinatorics, but how for ex AD is in alphabetical order?
I thought that we are restricted to use a pair of letters only in alphabetical order, ex: AB, BC, CD, etc.
that is my understanding of the alphabetical order.