PS - combination

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PS - combination

by abhasjha » Sun Nov 16, 2014 1:17 pm

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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

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by [email protected] » Sun Nov 16, 2014 2:13 pm
abhasjha 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.

ASIDE: If anyone is interested, we have a free video on calculating combinations (like 5C2) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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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by [email protected] » Sun Nov 16, 2014 2:17 pm
abhasjha 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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Brent
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by Mathsbuddy » Wed Nov 19, 2014 5:44 am
Imagine a 2 digit number AB where A and B are integers >= 0
Let's see what is the lowest base that can generate the value twelve using just these 2 digits?
Using base 4, the maximum number achievable using 2 digits is 33 (base 4) = 3x4 + 3 = 15 (denary)
However we need to remove 3 doubles (where A=B): 11,22,33 which leaves us with 15-3 = 12
and also remove half of these (as AB is accepted, but not BA), leaving us with 6
We can also add 4 for the 4 single digits (0,1,2,3)
Therefore Base 4 produces 10 possibilities
Hence any greater base will suffice (as the next base will produce plenty more)
Therefore base 5 is the lowest base.
Answer = B) 5.

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by Mathsbuddy » Wed Nov 19, 2014 6:57 am
Mathsbuddy wrote:Imagine a 2 digit number AB where A and B are integers >= 0
Let's see what is the lowest base that can generate the value twelve using just these 2 digits?
Using base 4, the maximum number achievable using 2 digits is 33 (base 4) = 3x4 + 3 = 15 (denary)
However we need to remove 3 doubles (where A=B): 11,22,33 which leaves us with 15-3 = 12
and also remove half of these (as AB is accepted, but not BA), leaving us with 6
We can also add 4 for the 4 single digits (0,1,2,3)
Therefore Base 4 produces 10 possibilities
Hence any greater base will suffice (as the next base will produce plenty more)
Therefore base 5 is the lowest base.
Answer = B) 5.
Using the same algorithm for base 5, we get:
Maximum number = 44 (base 5) = 4 x 5 + 4 = 24 (denary)
Remove 4 doubles (11, 22, 33, 44) -> 24 - 4 = 20
Half this to remove alphabetical reversals -> 10
Add 5 for the 5 single digits -> 10 + 5 = 15
15 > 12, so base 5 works
Answer = B) 5

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by GMATinsight » Thu Nov 27, 2014 10:23 am
abhasjha 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
A typical P&C approach

If n are the total letters then
total combination of 1 Letter code = n
total combination of 2 Letter code (in alphabetical order) = nC2 = n(n-1)/2

Therefore total combination = n + n(n-1)/2 which must be greater than or equal to 12
i.e. 2n+n^2-n >or= 24
i.e. n^2 +n >or= 24

i.e. Minimum value of n must be 5

Answer: Option B
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by Gurpreet singh » Sun Jun 26, 2016 11:17 pm
Brute force

1 A
2 B
3 AB
4 C
5 CA
6 CB
7 D
8 AD
9 DB
10 DC
11 E
12 EA

Alphabets used 5

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by [email protected] » Fri May 18, 2018 11:28 am
abhasjha 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 solve this problem by using combinations. We can let n denote the total number of letters needed. Since there are 12 participants and we have the option of one-letter codes and two-letter codes, we need the sum of nC1 and nC2 to be at least 12. That is:

nC1 + nC2 ≥ 12

From here, the easiest way to solve this problem is to test the answer choices. We can start with answer choice A.

A. 4

4C1 + 4C2 = 4 + (4x3)/2 = 4 + 2x3 = 10

If we use 4 letters, we will obtain only ten unique codes; this is insufficient for 12 participants.
B. 5

5C1 + 5C2 = 5 + (5x4)/2 = 5 + 5x2 = 15

We obtain 15 unique codes by using 5 letters, and this is more than sufficient for 12 participants.

Answer: B

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Re: PS - combination

by [email protected] » Sat Apr 24, 2021 7:49 pm
Hi All,

We’re told that each participant in an experiment will be referred to with a ‘code’ that is either 1 letter or 2 DISTINCT letters written in ALPHABETIC ORDER. We’re asked for the LEAST number of letters that would be needed to give each of 12 participants a unique code.

Since the answers are numbers, we can TEST THE ANSWERS solve this problem.

IF… there were 4 letters, then we can refer to those letters as A, B, C and D – and the possible codes would be:

A, B, C and D = 4 one-letter codes
AB, AC, AD, BC, BD, CD = 6 two-letter codes
Total = 10 codes… which is close, but too few.

Adding 1 additional letter will clearly create more than 12 codes, so you can stop working at this point. If you choose to list out the codes though, then we can refer to the letters as A, B, C, D and E – and the codes would be:

A, B, C, D and E = 5 one-letter codes
AB, AC, AD, AE, BC, BD, BE, CD, CE and DE = 10 two-letter codes
Total = 15 codes… which is more than we technically need, but the least number of total letters that will give us 12 total codes.

Final Answer: B

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