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# A researcher plans to identify each participant in a certain

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

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17 Jun 2012, 03:13
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 17 Jun 2012, 03:20, edited 1 time in total.
Edited the question.
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17 Jun 2012, 03:24
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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.
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17 Jun 2012, 03:34
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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

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

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14 Mar 2014, 01:48
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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 [#permalink]

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27 Dec 2014, 10:22
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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

[Reveal] Spoiler:
B

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

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02 Dec 2012, 13:48
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Hi Bunnel

won't this $$C^2_n$$
just give you all the pairs available?
we need them also ordered....
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Re: A researcher plans to identify each participant in a certain [#permalink]

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23 Dec 2012, 23:41
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eaakbari 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.

Bunuel,

What if the question didn't say 'pair'.

If 3 letter combinations were also permitted. How would you express it in Combination formula?

Practice: try to use the same concept.
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24 Dec 2012, 00:49
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eaakbari wrote:
Bunuel wrote:

Practice: try to use the same concept.

Okay here goes,

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); $$nC2$$
The # of Triples of distinct letters codes possible would be (in alphabetical order); $$nC3$$

Thus

$$nC3 + nC2 + n$$> $$12$$

$$n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n$$> $$12$$

Simplifying

$$n*(n^2 +5)$$> $$72$$

Only sufficient value of $$n = 4$$

Is it correct?

Correct.

Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.

Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.
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17 Jun 2012, 03:49
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Almost identical question:

John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
A. 24
B. 12
C. 7
D. 6
E. 5

The concept is not that hard. We can use combination or trial and error approach.

Combination approach:
Let # of colors needed be $$n$$, then it must be true that $$n+C^2_n\geq{12}$$ ($$C^2_n$$ - # of ways to choose the pair of different colors from $$n$$ colors when order doesn't matter) --> $$n+\frac{n(n-1)}{2}\geq{12}$$ --> $$2n+n(n-1)\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> as $$n$$ is an integer (it represents # of colors) $$n\geq{5}$$ --> $$n_{min}=5$$.

Trial and error approach:
If the minimum number of colors needed is 4 then there are 4 single color codes possible PLUS $$C^2_4=6$$ two-color codes --> 4+6=10<12 --> not enough for 12 codes;

If the minimum number of colors needed is 5 then there are 5 single color codes possible PLUS $$C^2_5=10$$ two-color codes --> 5+10=15>12 --> more than enough for 12 codes.

Actually as the least answer choice is 5 then if you tried it first you'd get the correct answer right away.

Hope it helps.
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03 Dec 2012, 01:30
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ronr34 wrote:
Hi Bunnel

won't this $$C^2_n$$
just give you all the pairs available?
we need them also ordered....

Notice that we are told that letters in the code should be written in alphabetical order. Now, 2Cn gives different pairs of 2 letters possible out of n letters, but since codes should be written in one particular order (alphabetical), then for each pair there will be only one ordering possible, thus the number of codes out of n letters equals to number of pairs out of n letters.

Hope it's clear.
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24 Dec 2012, 00:28
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Bunuel wrote:

Practice: try to use the same concept.

Okay here goes,

The # of single letter codes possible would be $$n$$ itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); $$nC2$$
The # of Triples of distinct letters codes possible would be (in alphabetical order); $$nC3$$

Thus

$$nC3 + nC2 + n$$> $$12$$

$$n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n$$> $$12$$

Simplifying

$$n*(n^2 +5)$$> $$72$$

Only sufficient value of $$n = 4$$

Is it correct?
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25 Nov 2013, 03:41
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honchos 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.

we can take 1,2 and 3
like
A, B, C
AB, BC
ABC

Why did you ignored possibility of 3 or 4 alphabets taken together, this will give us 4 letters?

Please read the question carefully: a code consists of either a single letter or a pair of distinct letters written in alphabetical order.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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29 Nov 2013, 19:18
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honchos 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.

we can take 1,2 and 3
like
A, B, C
AB, BC
ABC

Why did you ignored possibility of 3 or 4 alphabets taken together, this will give us 4 letters?

The question specifically points out that the combinations can be a 1 digit letter or a 2 digit letter.
I used a simple combination as stated in other answers to find out.

1. A
2. B
3. BA
4. C
5. CA
6. CB
7. D
8. DA
9. DB
10. DC
11. E
12. EA

STOP. you get the answer as 5 (ABCDE)
Also what i have found is that when writing down the combinations with no repeats, it is easier to start with one letter and keep repeating it until you exhausted all the options. this will eliminate confusion. like you start with C and repeat with CA CB and then with D DA DB DC..
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Re: A researcher plans to identify each participant in a certain [#permalink]

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13 Aug 2014, 23:54
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ccyang24 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.

Thanks Bunuel for the explanation.
I do need some clarification regarding the C(n,r).
How does C(n, 2) = n(n-1)/2?
Shouldn't it be n! / (2! (n - 2)!)

Thank you!

Hope it helps.
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10 Sep 2014, 11:21
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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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Re: A researcher plans to identify each participant in a certain [#permalink]

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09 Feb 2015, 00:43
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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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A researcher plans to identify each participant in a certain [#permalink]

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17 Aug 2016, 09:36
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g3lo18 wrote:
In addition to the above query, I am still unclear as to why we use combination formula to solve this particular question when order clearly matters? I read Bunuels response as to why, but I am still unclear. I must have spent 5 days reading the explanation.

Posted from my mobile device

(replying here instead of in chat as there's an MBA session going on).

The only thing that matters is the number of distinct pairs.

Stating that the order must be alphabetical states that only 1 of all potential pairs is valid. This is equivalent to the combination formula as shown below.

For example, with $$\{A,B,C,D\}$$, the list of permutations of size 2 is $$4_P 2 = 12$$

AB, BA = 2! ways to arrange 2 elements
AC, CA
BC, CB
BD, DB
CD, DC

Saying that the order of the selection does not matter is equivalent to saying that every permutation is the same which is equivalent to saying that only one of each permutation is valid.

Permutations = $$\frac{n!}{(n-k)!}$$
Combinations = $$\frac{n!}{k!(n-k)!}$$
Number of ways of arranging each new selection of elements = $$k!$$
Where 1 out of every new selection is valid = $$\frac{1}{k!}$$
Permutations where only 1 permutation is valid = $$\frac{n!}{(n-k)!} \times \frac{1}{k!} =$$ Combinations
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Re: A researcher plans to identify each participant in a certain [#permalink]

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01 Dec 2012, 19:10
Bunnel. Thaks for the reply and merging similar topics. Can u please explain how >= 12 ?
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Re: A researcher plans to identify each participant in a certain [#permalink]

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02 Dec 2012, 03:51
SreeViji wrote:
Bunnel. Thaks for the reply and merging similar topics. Can u please explain how >= 12 ?

The number of letters should be enough to make at least 12 codes, thus the number of codes must be more than or equal to 12.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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23 Dec 2012, 12:22
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.

Bunuel,

What if the question didn't say 'pair'.

If 3 letter combinations were also permitted. How would you express it in Combination formula?
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Re: A researcher plans to identify each participant in a certain   [#permalink] 23 Dec 2012, 12:22

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