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# A researcher plans to identify each participant in a certain

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

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25 Feb 2014, 08:57
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 $$n_{min}=5$$

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

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25 Feb 2014, 09:39
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 [#permalink]

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13 Mar 2014, 15:03
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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 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)! ?

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

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14 Mar 2014, 02:48
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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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06 Apr 2014, 12:38
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.

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

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06 Apr 2014, 13:09
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 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 $$n(n+1)\geq{0}$$.

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 [#permalink]

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18 May 2014, 00:24
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.

Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ?

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

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18 May 2014, 01:07
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gauravsaxena21 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.

Every time I see your explanation, problem becomes so easy, but when I tried my own, I hardly get the correct. How to improve my understanding on combination and Probability ?

By studying theory and practicing.

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Theory on probability problems: math-probability-87244.html

All DS probability problems to practice: search.php?search_id=tag&tag_id=33
All PS probability problems to practice: search.php?search_id=tag&tag_id=54

Tough probability questions: hardest-area-questions-probability-and-combinations-101361.html

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

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02 Sep 2014, 18:53
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.?

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

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02 Sep 2014, 20:16
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 [#permalink]

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10 Sep 2014, 11:41
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.

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

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10 Sep 2014, 11:54
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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 [#permalink]

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10 Sep 2014, 12:06
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.

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

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10 Sep 2014, 12:21
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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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22 Nov 2014, 01:47
Hi Bunuel,
if the same question leaves out the requirement of arranging with alphabetical order, the answer will be
n+ n(n-1) > 12
n2>12 => n= 4
Is this correct?

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

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22 Nov 2014, 06:44
vietnammba wrote:
Hi Bunuel,
if the same question leaves out the requirement of arranging with alphabetical order, the answer will be
n+ n(n-1) > 12
n2>12 => n= 4
Is this correct?

Yes.

Check other similar questions: Constructing Numbers, Codes and Passwords.
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A researcher plans to identify each participant in a certain [#permalink]

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14 Dec 2014, 06:58
My solution was based on answers:
(A) 4 Letters: 4 Letters + 4!/2!*2! = 10 --> So it is not enough (2! means when out of 4 letters there are 2 letters in the code and 2 letters aren't)
(B) 5 Letters: 5 Letters + 5!/3!*2! = 15 --> So it's enough to build a code for 12 Persons

Solving with factorials, I calculated the amount of combinations of 2 distinct Letters for 4 or 5 Letters, so by 5 Letters, if 2 letters are in the code 3 of them are not --> 2! ans 3!.

I hope my solution was clear enough for you
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Re: A researcher plans to identify each participant in a certain [#permalink]

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26 Dec 2014, 06:40
I did this:

I also started by thinking than 12 single letters give 12 diffferent codes.
So, to use less, 12/2= 6 letters, that combined can give 12 codes (or more).

Then, I just created a letter matrix like this, using only the 6 first letters of the alphabet, and started selecting:
A B C D E F --> I 1st created the first row (6 letters max) and recreated the sequence beneath each letter
A A A A A A
B B B B B B
C C C C C C
D D D D D D
E E E E E E
F F F F F F

Then, I started selecting, using the least number of letters possible:
I selected the 1st row A, then the B beneath it --> A, AB (I deleted the second single A, as it is not allowed; same for rest)
I selected the 1st row B, then C beneath it --> B, BC
I selected the 1st row C, then the A beneath it (not to use more letters) --> C, AC
I selected 1st row D, then A,B,C beneath it (again using as much as the previous) --> D, AD, BD,CD
I selected 1st row E, then A beneath it (for the last 2 digits) --> E, AE.

Now I have 12 numbers, consisting either of one or two digits, in alphabetical order. I used 5 letters.

It can happen quite quickly, because having the matrix down on paper you can just cross out or select letters.

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

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27 Dec 2014, 11:22
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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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08 Feb 2015, 09:44
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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