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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 [#permalink]

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26 Feb 2015, 07:06

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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visit my facebook to help me. on facebook, my name is: thang thang thang

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.

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

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29 Apr 2015, 01:10

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 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?

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.

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

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21 Nov 2015, 17:55

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.

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:

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

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04 Mar 2016, 10:39

Bunuel wrote:

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.

Hi Bunuel,

From n letters we choose the number of pairs, the result will be \(C^2_n\) which may include 2 kinds of pairs (AB) and (BA). Still confused .[/quote]

Maybe the following example would help. Consider 4 letters {a, b, c, d}. How many 2-letter words in alphabetical order are possible? The answer is \(C^2_4=6\): ab; ac; ad; bc; bd; cd.[/quote]

So it is essentially COMBINATION that matters... even if it were not told alphabetical, then we had to consider permutation... so AB and BA would be different !! Got it now.. Thanks Bunuel... some where in the wording of the Qs , it seemed like we needed to do a nc2 / 2.....

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

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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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17 Aug 2016, 10:36

1

This post received KUDOS

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

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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?

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

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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: \(\{A,B\})\).

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

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03 Sep 2016, 11:46

here is how I did this. let us say we have 4 to begin with. A,B,C,D we can have A,B,C,D ==> 4 AB, AC, AD - 3 with first letter as A 2 with first letter as B 1 with first letter as C. so in total we have 4+3+2+1= 2.5*4 = 10. Notice that this is as Bunuel explained is nC2+n.

By this we can apply intuition and see that with 5 we will definitely climb 12. nC2 + 5 = 10+5 = 15. We only needed 12. So B.

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

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21 Mar 2017, 12:46

Since the question says "either one or a pair of distinct letter", i made the calculation as in ; " A, AB,BA, B, C,CB,CA,... etc therefore 4 letters were sufficient. How am i supposed to understand that if we are going to use "only single letters or pair of distinct letters" alltogether?

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

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25 Mar 2017, 19:16

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);