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# A researcher plans to identify each participant in a certain

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Math Expert
Joined: 02 Sep 2009
Posts: 44413
Re: A researcher plans to identify each participant in a certain [#permalink]

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26 Mar 2017, 04:57
Avinash_R1 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.

in this problem
If i write the sequence - A, B, C, D, AB, AC, AD, BA, BC, BD, CA, CB
why we should not consider AB and BA as separate entity?

We are told ion the stem that the codes must be in alphabetical order. BA is not in alphabetical order. BTW this is explained many times on previous pages.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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24 Apr 2017, 02:15
Bunuel wrote:
russ9 wrote:
Hi Bunuel,

Thanks for the clarification. I was having a hard time grasping the equation itself but I followed a link to the mathbook topic and that does a good job of explaining why the equation is the way it is.

What I do question is the arrangement of the letters. I did go to the two links you posted and it's still a little unclear. How does the 2Cn equation know to not double count BA and AB. Wouldn't we have to go to the permutation equation for that?

Apart from that link I can only advice you to check it yourself. How many 2-letter words in alphabetical order are possible from say 3 letters {a, b, c}.

Just saw this! Thanks Bunuel!

I only found 1 possible order to {a, b, c} in alphabetical order, but if it works that way then I am wondering what is the use of Permutations in that case??
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Re: A researcher plans to identify each participant in a certain [#permalink]

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02 May 2017, 21:09
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

My 2 cents.
As I am not good at combination, I just started backsolving.

A. 4.

Ok so 4 letters can be...(caution : it has to be in alphabetical order)

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

Ok...so 4 is not enough...
Well, lets add one more letter...

11. E
12. AE...
.....
...

So, we need at least 5.
Hence B.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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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 $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

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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Joined: 02 Sep 2009
Posts: 44413
Re: A researcher plans to identify each participant in a certain [#permalink]

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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 $$C^2_n+n\geq{12}$$ --> $$\frac{n(n-1)}{2}+n\geq{12}$$ --> $$n(n-1)+2n\geq{24}$$ --> $$n(n+1)\geq{24}$$ --> $$n_{min}=5$$.

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 [#permalink]

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08 Jun 2017, 01:53
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

Pair of distinct letters
Hence, If we take
4!/2! , we will get 12 ways to arrange it, but we will have to divide by 2 to minus the repetitions which will give us 6 ways.
We would have to take minimum 5 letters to satisfy condition.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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17 Jun 2017, 09:59
The way I understand =>
Reverse Combination question - Normally we have to find number of ways to select. Here we are given the final selection and need to find least objects to be used. We have either 1 seat or 2 seats to fill. This can be done in r or nCr ways

So to create 12 codes from minimum letter we need 5 + 5C2 = 5 + 10 = 15.
Here Arrangement is not considered as we are not allowed arrangement like BA, DA.
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A researcher plans to identify each participant in a certain [#permalink]

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03 Sep 2017, 11:27
Take A and B first.
Then see the possible code combinations. The combinations are A, B, AB, BA

Similarly, take B and C. Possible combinations are B, BC and CB

In this way, the 11th code will be DC and 12th code will be E.

So least number of alphabets used are A,B,C,D and E,

The answer is 5 (option B)
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Re: A researcher plans to identify each participant in a certain [#permalink]

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25 Sep 2017, 21:36
Trail and Error is the best method,
A
B
C
D
AB
AC
BC
BD
CD
only ten possible as shown above if we take 4 letters,
So answer should be 5 letters
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Re: A researcher plans to identify each participant in a certain [#permalink]

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

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

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17 Oct 2017, 12:09
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!

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

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28 Nov 2017, 19:19
@Brunuel,

I was also confused as to how you derived $$C^2_n=\frac{n!}{2!(n-2)!}$$ into$$\frac{(n-1)n}{2}$$

I'm only used to using the "standard" form $$\frac{n!}{2!(n-2)!}$$. Is there a general rule I can/should learn from this?

Thanks so much,

Bunuel wrote:

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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28 Nov 2017, 20:09
@Brunuel,

I was also confused as to how you derived $$C^2_n=\frac{n!}{2!(n-2)!}$$ into$$\frac{(n-1)n}{2}$$

I'm only used to using the "standard" form $$\frac{n!}{2!(n-2)!}$$. Is there a general rule I can/should learn from this?

Thanks so much,

Bunuel wrote:

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.

This is already explained here: https://gmatclub.com/forum/a-researcher ... l#p1344122

$$C^2_n=\frac{n!}{2!(n-2)!}$$. Now, notice that $$n!=(n-2)!*(n-1)*n$$, hence $$C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}$$.
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A researcher plans to identify each participant in a certain [#permalink]

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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.

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

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

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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.

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 [#permalink]

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19 Dec 2017, 09:28
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.

Hello Bunuel, the combination formula is N! / ( N-R)! R! , my question is why are why are you writing (n-1) n - so I simply trying to understand logic behind what you wrote. I visited this link https://gmatclub.com/forum/math-combina ... 87345.html but could not find answer why you put (n-1)n in the numerator, whereas as per formula (n-1)n is in the denominator, and why you divide by 2 ? Thanks a lot for taking time to explain!, D.
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Re: A researcher plans to identify each participant in a certain [#permalink]

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19 Dec 2017, 09:46
dave13 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.

Hello Bunuel, the combination formula is N! / ( N-R)! R! , my question is why are why are you writing (n-1) n - so I simply trying to understand logic behind what you wrote. I visited this link https://gmatclub.com/forum/math-combina ... 87345.html but could not find answer why you put (n-1)n in the numerator, whereas as per formula (n-1)n is in the denominator, and why you divide by 2 ? Thanks a lot for taking time to explain!, D.

This is already explained here: https://gmatclub.com/forum/a-researcher ... l#p1344122

$$C^2_n=\frac{n!}{2!(n-2)!}$$. Now, notice that $$n!=(n-2)!*(n-1)*n$$, hence $$C^2_n=\frac{n!}{2!(n-2)!}=\frac{(n-2)!*(n-1)*n}{2!(n-2)!}=\frac{(n-1)n}{2}$$.

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

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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.

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

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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
Re: A researcher plans to identify each participant in a certain   [#permalink] 19 Dec 2017, 15:26

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