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Math Expert V
Joined: 02 Sep 2009
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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amz14 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

I have a questions here:
How did we get from $$n(n+1)\geq{24}$$ to $$n_{min}=5$$

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

Hence, $$n_{min}=5$$.

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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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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2
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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)! ?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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3
4
RebekaMo wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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russ9 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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 a combination of 2 letters out of a group of n letters should 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?

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 and understand the content of the guide but these two topics still elude me. I'm open to learning from another venue if helpful?

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

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Hope this helps.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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.?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Chin926926 wrote:
Bunuel wrote:
sarb wrote:
A researcher plans to identify each participant in a certain medical experiment with a code consisting of either a single letter or a pair of distinct letters written in alphabetical order. What is the least number of letters that can be used if there are 12 participants, and each participant is to receive a different code?

A. 4
B. 5
C. 6
D. 7
E. 8

Say there are minimum of $$n$$ letters needed, then;

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be $$C^2_n$$ (in alphabetical order);

We want $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

Hope it's clear.

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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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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1
1
kevn1115 wrote:
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.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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1
kevn1115 wrote:
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.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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5
2
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

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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Bunuel wrote:
kevn1115 wrote:
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.

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!
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Math Expert V
Joined: 02 Sep 2009
Posts: 53831
Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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1
ProblemChild wrote:
Bunuel wrote:
kevn1115 wrote:
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.

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???
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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1
you can not remember in the test room

think about the basic problem in Princeton gmat book.

if there is 2 element, we have 2x1 possibilites , order maters
if there is 3 element, we have 3x2
if there is 5 element, we have 5x4
but because the element must be in alphatical order, we have half of the possibilities
5x4/2=10
we have 5 distick letter, so total is 15
enough
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Hi Folks,

Pair would mean 2 elements.So we will consider single and pairs(with 2 elements) only right?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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kirtivardhan wrote:
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
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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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?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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kirtivardhan wrote:
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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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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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. Re: A researcher plans to identify each participant in a certain medical   [#permalink] 21 Nov 2015, 17:55

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# A researcher plans to identify each participant in a certain medical

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