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A researcher plans to identify each participant in a certain

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New post 22 Nov 2015, 00:13
Hi mvictor,

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
AD
BC
BD
CD

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New post 25 May 2016, 18:47
Attached is a visual that should help.
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Screen Shot 2016-05-25 at 6.44.12 PM.png
Screen Shot 2016-05-25 at 6.44.12 PM.png [ 64.3 KiB | Viewed 1018 times ]


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New post 15 Jun 2016, 06:06
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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.

Answer B
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New post 17 Aug 2016, 08:45
Hi,

Could someone please explain how we went from n!/2!(n-2)! -> n(n-1)/2 ?

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.

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New post 17 Aug 2016, 10:36
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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.

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(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
AD, DA
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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New post 17 Aug 2016, 10:53
Thank you for that.

I guess my confusion stems from watching the GMATPrepNow videos. Brent explains that if the order does not matter, we use nCr formula. However, in this case it does matter (alphabetical). I guess there may be an error in his videos?
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New post 17 Aug 2016, 11:22
g3lo18 wrote:
Thank you for that.

I guess my confusion stems from watching the GMATPrepNow videos. Brent explains that if the order does not matter, we use nCr formula. However, in this case it does matter (alphabetical). I guess there may be an error in his videos?


Sorry, this is relatively hard to explain and I don't feel I'm making a good job of explaining it.

In this question, the order does not matter.

The problem lies with the interpretation of the phrase "the order matters". Specifically it means: when selecting a set of elements for the output, every permutation of that set is valid. This means that \(\{A,B,D\}\) and \(\{D,A,B\}\) are both distinct and valid elements in the solution set.

A combination on the other hand specifies that \(\{A,B,D\}\) and \(\{D,A,B\}\) are equivalent (which can also be described by saying that the order does not matter). This means that for the set of possible solutions containing three elements, only one of which is valid.

In this question, every pair of letters maps to a single solution: out of \(\{\{A,B\},\{B,A\}\}\), only \(\{A,B\}\) is valid. We are performing the operation of moving from a set of elements to a single element. Therefore we use combinations.

EDIT (may be more clear): In this question, we are first taking a collection of letters and determining how many pairs of distinct letters we can make (permutations). For every pair of distinct letters, we are then mapping from that collection to a single element. This map from a collection to a single element is a combination (all elements in the set of \(\{\{A,B\},\{B,A\}\}\) are not a valid solutions, but each is a map to the single valid solution in the set: \(\{A,B\})\).

There is a good video explaining the concept here (the example at 5:01): https://www.coursera.org/learn/fe-exam/ ... mbinations

I hope this helps (and sorry that I can't explain things better).
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New post 10 May 2017, 06:33
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 \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

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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New post 10 May 2017, 06:37
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 \(C^2_n+n\geq{12}\) --> \(\frac{n(n-1)}{2}+n\geq{12}\) --> \(n(n-1)+2n\geq{24}\) --> \(n(n+1)\geq{24}\) --> \(n_{min}=5\).

Answer: B.

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: \(C^2_n=\frac{n!}{2!(n-2)!}\)

Combinatorics Made Easy!

Theory on Combinations

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New post 03 Oct 2017, 09:28
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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New post 03 Oct 2017, 09:48
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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New post Updated on: 29 Nov 2017, 19:47
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Since the question asks for the minimum number of letters, it makes sense to start with the least answer (choice A) and work your way down.

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!

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Originally posted by mcelroytutoring on 17 Oct 2017, 12:09.
Last edited by mcelroytutoring on 29 Nov 2017, 19:47, edited 1 time in total.
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New post 06 Dec 2017, 09:05
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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


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.

Answer choice A (4 letters)
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.

Answer choice B (5 letters)
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

Answer: B

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New post 06 Dec 2017, 09:08
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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


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

Answer choice A (4 letters)
Let the letters be A, B, C, D
The possible codes are:
A
B
C
D
AB
AC
AD
BC
BD
CD
TOTAL = 10 (not enough. We need at least 12 codes)

Answer choice B (5 letters)
Let the letters be A, B, C, D, E
The possible codes are:
A
B
C
D
E
AB
AC
AD
AE
BC
BD
BE
CD
CE
DC
TOTAL = 15

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

Answer: B

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New post 19 Dec 2017, 11:39
Let the first spot be A. We have to use another letter for the next slot so let it be B. For the next we have to ask ourselves can we make an entry without introducing another letter? we can. so far we have
A, B, AB,____. For the fourth space must we introduce a letter? the answer is yes.

so we have A, B, AB, C, ____,______

for the fifth and sixth space we can have AC and BC

A, B, AB, C, AC, BC,____,______,______

We must introduce another letter so let it be D

A, B, AB, C, AC, BC, D, AD, BD, CD, ____,_____

We have to introduce another letter so let it be E.

A, B, AB, C, AC, BC, D, AD, BD, CD, E, AE, BE, CE, DE

We can go up to fifteen. We need 12 so the total different letters we used were A, B, C, D, and E.

The answer is 5.
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New post 19 Dec 2017, 15:26
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 I`ve 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 :? :)
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New post 19 Dec 2017, 20:36
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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 I`ve 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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A researcher plans to identify each participant in a certain  [#permalink]

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New post 09 Jan 2018, 08:12
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 use just logic to solve this question.

Here we are given that all letters and we need to get the least number of letters for 12 participants and all the letter should in alphabetical order.

Let's choose 4 letters since this is the least number.
A B C D
Then we can choose two letters again , AB, AC, AD, BC, BD, BD - total 6 ways. ( Note here we can't BA since we need to follow alphabetical order )
Total 4 + 6 = 10 and this suits only for 10 participants.

Let's choose 5 letters.
A B C D E - 5 ways
AB AC AD AE BC BD BE CD CE DE - 10 ways . We got 15 ways and 15 different codes are sufficient to give the code for 12 participants.

Hence 5.
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A researcher plans to identify each participant in a certain &nbs [#permalink] 09 Jan 2018, 08:12

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