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Difficulty:
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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
A. 4
B. 5
C. 6
D. 7
E. 8
sarb
Say there are minimum of \(n\) letters needed, then;
- The number of single-letter codes possible would be \(n\) itself;
- The number of two-letter codes possible (pair of distinct letters) 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.
Or else one could just test numbers.
- Four letters are enough for \(4+C^2_4=4+6=10\) codes. Not enough.
- Five letters are enough for \(5+C^2_5=5+10=15\) codes. Good!
Hope it's clear.
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sarb
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Almost identical question:
John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
A. 24
B. 12
C. 7
D. 6
E. 5
The concept is not that hard. We can use combination or trial and error approach.
Combination approach:
Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).
Trial and error approach:
If the minimum number of colors needed is 4 then there are 4 single color codes possible PLUS \(C^2_4=6\) two-color codes --> 4+6=10<12 --> not enough for 12 codes;
If the minimum number of colors needed is 5 then there are 5 single color codes possible PLUS \(C^2_5=10\) two-color codes --> 5+10=15>12 --> more than enough for 12 codes.
Actually as the least answer choice is 5 then if you tried it first you'd get the correct answer right away.
Answer: E.
Hope it helps.
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John has 12 clients and he wants to use color coding to identify each client. If either a single color or a pair of two different colors can represent a client code, what is the minimum number of colors needed for the coding? Assume that changing the color order within a pair does not produce different codes.
A. 24
B. 12
C. 7
D. 6
E. 5
The concept is not that hard. We can use combination or trial and error approach.
Combination approach:
Let # of colors needed be \(n\), then it must be true that \(n+C^2_n\geq{12}\) (\(C^2_n\) - # of ways to choose the pair of different colors from \(n\) colors when order doesn't matter) --> \(n+\frac{n(n-1)}{2}\geq{12}\) --> \(2n+n(n-1)\geq{24}\) --> \(n(n+1)\geq{24}\) --> as \(n\) is an integer (it represents # of colors) \(n\geq{5}\) --> \(n_{min}=5\).
Trial and error approach:
If the minimum number of colors needed is 4 then there are 4 single color codes possible PLUS \(C^2_4=6\) two-color codes --> 4+6=10<12 --> not enough for 12 codes;
If the minimum number of colors needed is 5 then there are 5 single color codes possible PLUS \(C^2_5=10\) two-color codes --> 5+10=15>12 --> more than enough for 12 codes.
Actually as the least answer choice is 5 then if you tried it first you'd get the correct answer right away.
Answer: E.
Hope it helps.
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EMPOWERgmatRichC
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Expert reply
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
Final Answer:
GMAT assassins aren't born, they're made,
Rich
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
Final Answer:
GMAT assassins aren't born, they're made,
Rich
ronr34
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.
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sarb
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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ScottTargetTestPrep
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Joined: 14 Oct 2015
Last visit: 09 Sep 2026
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Expert reply
Posts: 22,736
sarb
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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RebekaMo
\(C^2_n=\frac{n!}{2!(n-2)!}\). 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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sarb
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
Cheers,
Brent
General Discussion
Bunnel. Thaks for the reply and merging similar topics. Can u please explain how >= 12 ?
SreeViji
The number of letters should be enough to make at least 12 codes, thus the number of codes must be more than or equal to 12.
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Hi Bunnel
won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....
won't this \(C^2_n\)
just give you all the pairs available?
we need them also ordered....
Bunuel
Bunuel,
What if the question didn't say 'pair'.
If 3 letter combinations were also permitted. How would you express it in Combination formula?
eaakbari
Practice: try to use the same concept.
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Bunuel
Okay here goes,
The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)
Thus
\(nC3 + nC2 + n\)> \(12\)
\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)> \(12\)
Simplifying
\(n*(n^2 +5)\)> \(72\)
Only sufficient value of \(n = 4\)
Is it correct?
eaakbari
Correct.
Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.
Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
AD;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.
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Bunuel
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
.
yenpham9
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.
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Lets take A B C D
A
B
C
D
AB
AC
AD
BC
BD
CD
ABC
BCA
CBA
It is alphabetical and all letter for a particular codes are different.
A
B
C
D
AB
AC
AD
BC
BD
CD
ABC
BCA
CBA
It is alphabetical and all letter for a particular codes are different.
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