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

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Attached is a visual that should help.
Attachments Screen Shot 2016-05-25 at 6.44.12 PM.png [ 64.3 KiB | Viewed 1825 times ]

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

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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

Let's use the answer choices to help us solve this problem. We are looking for the minimum number of letters that can be used. The smallest number from the answer choices is 4, so let’s ask ourselves this question: Can we use only 4 letters to represent the 12 participants? Assume that the 4 letters are A, B, C and D (keep in mind that for each participant we can use either one letter OR two letters to represent him or her; however if we use two letters, they must be in alphabetical order):

1) A 2) B 3) C 4) D 5) AB 6) AC 7) AD 8) BC 9) BD 10) CD

Under this scheme, we can represent only 10 of the 12 participants. So let's add in one more letter, say E, and see if having an additional letter allows us to have a unique identifier for each of the 12 participants:

1) A 2) B 3) C 4) D 5) AB 6) AC 7) AD 8) BC 9) BD 10) CD 11) E 12) AE

As you can see, once we’ve added in the letter E we can represent all 12 participants. Since we’ve used A, B, C, D and E, the minimum number of letters that can be used is 5.

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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.

Can someone explain what this $$C^2_n+n\geq{12}$$ means? I also saw a reference to the same type of symbol with an A instead, what does that mean?
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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brandon7 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.

Can someone explain what this $$C^2_n+n\geq{12}$$ means? I also saw a reference to the same type of symbol with an A instead, what does that mean?

C stands for combinations: $$C^2_n=\frac{n!}{2!(n-2)!}$$

Theory on Combinations

DS questions on Combinations
PS questions on Combinations

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

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Hi Bunuel,

Can you please share the number of codes possible in alphabets. I am still confused with the combinations between 5 alphabets and assigning 12 unique codes with two distinct letters. Thanks.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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AnubhavK wrote:
Hi Bunuel,

Can you please share the number of codes possible in alphabets. I am still confused with the combinations between 5 alphabets and assigning 12 unique codes with two distinct letters. Thanks.

The number of codes consisting of either a single letter or a pair of distinct letters from 26-letter alphabet is $$26+C^2_{26}$$.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Top Contributor

A) 4 choose 1 is 4, and 4 choose 2 is 6. Unfortunately this only adds up to 10, and 10 < 12.
B) 5 choose 1 is 5 and 5 choose 2 is 10. This adds up to 15, and 15 > 12. We have a winner!

-Brian

Originally posted by mcelroytutoring on 17 Oct 2017, 11:09.
Last edited by mcelroytutoring on 29 Nov 2017, 18:47, edited 1 time in total.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Top Contributor
1
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

One approach is to add a BLANK to the letters in order to account for the possibility of using just one letter for a code.

ASIDE: Notice that, if we select 2 characters, there's only 1 possible code that can be created. The reason for this is that the 2 characters must be in ALPHABETICAL order. Or, in the case that a letter and a blank are selected, there's only one possible code as well.

Now we'll test the answer choices.

Let the letters be A, B, C, D
We'll add a "-" to represent a BLANK.
So, we must choose 2 characters from {A, B, C, D, -}
In how many ways can we select 2 characters?
We can use combinations to answer this. There are 5 characters, and we must select 2. This can be accomplished in 5C2 ways (= 10 ways).
So, there are only 10 possible codes if we use 4 letters. We want at least 12 codes.

[i]ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head below.

Let the letters be A, B, C, D, E
Once again, we'll add a "-" to represent a BLANK.
So, we must choose 2 characters from {A, B, C, D, E, -}
There are 6 characters, and we must select 2. This can be accomplished in 6C2 ways (= 15 ways...PERFECT).

So, the least number of characters needed is 5

RELATED VIDEO

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

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Top Contributor
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

We can also TEST each answer choice by LISTING all possible codes.

Let the letters be A, B, C, D
The possible codes are:
A
B
C
D
AB
AC
BC
BD
CD
TOTAL = 10 (not enough. We need at least 12 codes)

Let the letters be A, B, C, D, E
The possible codes are:
A
B
C
D
E
AB
AC
AE
BC
BD
BE
CD
CE
DC
TOTAL = 15

Perfect, 5 letters will give us the 12 codes we need.

Cheers,
Brent
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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.

Bunuel -Thank you ! Now when Ive reviewed the whole thread and still trying to understand some moments - why do you write it in this order 1*2*...*(n-4)*(n-3)(n-2)(n-1)n and not vice versa like this 1*2*...n(n-1)(n-2)(n-3)(n-4) etc ...also why you say" notice that n!=(n-2)(n-1)n" yes it as n! is in numerator as per formula and unlike formula, you simplify n! = n!/2!(n-2)! into != (n-2)(n-1)n" / 2!(n-2)!
first cant not "notice" the important detail you are trying to imply by pointing at this ---> n!=(n-2)(n-1)n - can I be helped with further explanation to understand this "notice" because in other combination formulas we didn't apply such simplification  Math Expert V
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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1
dave13 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.

Bunuel -Thank you ! Now when Ive reviewed the whole thread and still trying to understand some moments - why do you write it in this order 1*2*...*(n-4)*(n-3)(n-2)(n-1)n and not vice versa like this 1*2*...n(n-1)(n-2)(n-3)(n-4) etc ...also why you say" notice that n!=(n-2)(n-1)n" yes it as n! is in numerator as per formula and unlike formula, you simplify n! = n!/2!(n-2)! into != (n-2)(n-1)n" / 2!(n-2)!
first cant not "notice" the important detail you are trying to imply by pointing at this ---> n!=(n-2)(n-1)n - can I be helped with further explanation to understand this "notice" because in other combination formulas we didn't apply such simplification  1. n! is the product of integers from 1 to n, inclusive. So, n! = 1*2*...*(n-4)(n-3)(n-2)(n-1)n (1 is the smallest and n is the largest). Yes, you can write this in any order but it does not change anything.

2. n! = (n-2)!*(n-1)*n because (n-2)! = 1*2*...*(n-4)(n-3)(n-2), so (n-2)!*(n-1)*n = [1*2*...*(n-4)(n-3)(n-2)](n-1)n = n!

3. We can write $$C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}$$ whenever it's necessary.

Hope it's clear now.
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GMAT 1: 740 Q50 V40 Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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The question says that the letters have to be in alphabetical order so in this case can we keep AC too.
Since while writing AC we are missing B and not writing in alphabetical order
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Hi hassu13,

There's a difference between alphabetical order and CONSECUTIVE alphabetical order (in the same way that there's a difference between putting integers in numerical order and dealing with consecutive integers).

As an example, when dealing with the letters A, B, C and D there are 6 different pairs of letters that you could put in alphabetical order:

AB
AC
BC
BD
CD

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

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Bunuel
so effectively we have used 4 letters.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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sacmenon wrote:
Bunuel
so effectively we have used 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 medical  [#permalink]

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1
Got tripped up a bit on the wording, but saw the familiarity of the problem. The hard part is that these combination/permutation word problems express things in an particular way that doesn't always make it immediately clear whether order matters or not.

The issue I had was with the use of the word order, which in my mind would typically be associated with permutations, but the restriction of the problem actually means it's a combination.

"written in alphabetical order" - meaning that the letters have to be alphabetically ordered, but not necessarily consecutively next to each other on the alphabet. This actually makes it a combination and not a permutation question, because it removes all the other possibilities (i.e. we can have AB but not BA). Drawing it out makes sense: The fact that the letters have to be ordered alphabetically is a convoluted way of saying only unique combinations of letters matter. Hopefully my thinking is correct in this regard.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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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

Bunuel, chetan2u

Can any one elaborate condition of highlighted part.
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Re: A researcher plans to identify each participant in a certain medical  [#permalink]

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Top Contributor
Gmatprep550 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

Bunuel, chetan2u

Can any one elaborate condition of highlighted part.

It means that the letters we choose must be presented in alphabetical order.
So, for example, EG is good, since E comes before G in the alphabet.
Conversely, GE is not acceptable, since G does not come before E in the alphabet.

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

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Bunuel, can you explain why this is a combination and not a permutation?

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

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Hi crushorange,

The last sentence in the prompt tells us: "Assume that changing the color order within a pair does NOT produce different codes." This means that two different colors will only produce ONE code. For example Blue-Green and Green-Blue are the SAME code. Thus, the order of the colors does NOT matter and we're dealing with a Combination.

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_________________ Re: A researcher plans to identify each participant in a certain medical   [#permalink] 29 May 2019, 18:32

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