M04-29

Question

John has 12 clients, and he wants to use color coding to identify each of them. He can use either a single color or a pair of two different colors to represent a client code. Assuming that switching the order of colors within a pair does not create a different code, what is the minimum number of colors needed for this coding scheme?

A. 24
B. 12
C. 7
D. 6
E. 5

Show Answer

Official Answer

E

A. 24
B. 12
C. 7
D. 6
E. 5

Bunuel · Accepted Answer

Official Solution:

John has 12 clients, and he wants to use color coding to identify each of them. He can use either a single color or a pair of two different colors to represent a client code. Assuming that switching the order of colors within a pair does not create a different code, what is the minimum number of colors needed for this coding scheme?

A. 24
B. 12
C. 7
D. 6
E. 5

Show Answer

Official Answer

E

A. 24
B. 12
C. 7
D. 6
E. 5

Combination approach: 
 
Let the number of colors needed be  n . Then, ensuring that the total number of unique color codes (single color codes + two-color codes) is sufficient for all 12 clients, the inequality  n + C^2_n \ge 12  must hold.
 
 n+\frac{n(n-1)}{2} \ge 12 ;
 
 2n+n(n-1) \ge 24 ;
 
 n(n+1) \ge 24 . As  n  is an integer (it represents the number of colors), then  n \ge 5 , so  n_{	ext{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. Total:  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. Total:  5+10=15 > 12 . More than enough for 12 codes.
 
As the least answer choice is 5, if you tried it first, you would arrive at the correct answer immediately.

Answer: E

Bunuel · Answer

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} . Similar questions to practice: a-company-that-ships-boxes-to-a-total-of-12-distribution-cen-168427.html each-student-at-a-certain-university-is-given-a-four-charact-151945.html all-of-the-stocks-on-the-over-the-counter-market-are-126630.html if-a-code-word-is-defined-to-be-a-sequence-of-different-126652.html a-4-letter-code-word-consists-of-letters-a-b-and-c-if-the-59065.html a-5-digit-code-consists-of-one-number-digit-chosen-from-132263.html a-company-that-ships-boxes-to-a-total-of-12-distribution-95946.html a-company-plans-to-assign-identification-numbers-to-its-empl-69248.html the-security-gate-at-a-storage-facility-requires-a-five-109932.html all-of-the-bonds-on-a-certain-exchange-are-designated-by-a-150820.html a-local-bank-that-has-15-branches-uses-a-two-digit-code-to-98109.html a-researcher-plans-to-identify-each-participant-in-a-certain-134584.html baker-s-dozen-128782-20.html#p1057502 in-a-certain-appliance-store-each-model-of-television-is-136646.html john-has-12-clients-and-he-wants-to-use-color-coding-to-iden-107307.html how-many-4-digit-even-numbers-do-not-use-any-digit-more-than-101874.html a-certain-stock-exchange-designates-each-stock-with-a-85831.html the-simplastic-language-has-only-2-unique-values-and-105845.html the-telephone-company-wants-to-add-an-area-code-composed-of-20252.html Check below post for more: Probability Made Easy! Theory on probability problems Data Sufficiency Questions on Probability Problem Solving Questions on Probability Tough Probability Questions

him1985 · Answer

Let minimum n colors are required
We can write it as

nC1 + nC2 = > 12

Then try to put values from middle
7C1 + 7C2 = > 7+21 = 29

lets try lower value, if we can get value near to 12

Lets take n=5
5C1 + 5C2 = 5 + 10 = 15

This matches our requirement. We do not have smaller value than 5. So 5 is the answer

rohitd80 · Answer

I wonder of this is a real GMAT question?

This question is asking for the min # of colors than can make >= 12 color combinations (single color + twin color).

So, lets start with 2 color code set:
If we have 4 colors: the total 2 color combo's will be 4X3=12. But its said that order doesn't matter. In other words red+green=green+red. So, 12/2 = 6 two color codes. What additional # of codes do we get by using the same 4 colors individually...it's 4. So total color code combinations = 6+4 =10......But we need 12.

So, doing the same analysis with with 5 colors, we get....(two color codes)  + (1 color codes)  = 10+5=15......Answer=5

scofield1521 · Answer

Why would you start from 2 when the minimum number in options is 5!

If 5 Simply make combinations as it wont take time to calculate (I struggle with formulas, my bad)
A B C D E
AB AC AD AE
BC BD BE
We are done.
5 is the answer.

Icecream87 · Answer

Thanks Bunuel. I think I really need to go through my fundamentals on this again.

Senthil7 · Answer

I think this is a  high-quality  question and I agree with explanation.

joe123x · Answer

start from E.

if we have 5 colors and we want to know how many 2 colors combination can give us, then 5C2 = 10. ok so best case scenario in this case would be if we have 10 two color code, hold on, we still have 5 individual colors that we can assign to the remaining 2 clients. so, we don't need to continue, 5 is the answer.

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

vedha0 · Answer

hey Bunuel! there is quite some nuance in this question which can make one easily overlook and land up in the wrong answer, which are close choices present in the options as well. so i think it falls more in 650-700 level question. doesn't it?

Bunuel · Answer

We do not assign the difficulty level manually. The difficulty level of a question on the site is determined automatically based on various parameters collected from users' attempts, such as the percentage of correct answers and the time taken to answer the question. You can find the difficulty level of a question and its related statistics in the first post.

vedha0 · Answer

oh okay. thanks for the clarification!

Ishita2000 · Answer

Doesn't - "He can use either a single color or a pair of two different colors to represent a client code." mean that either 1 color - i.e. 12 colors or nC2=12...not a combination of both?

Bunuel · Answer

No, that means the code can consist of either one color OR two colors.

ddvij · Answer

I like the solution - it’s helpful.

bondi123 · Answer

This is a great question that’s helpful for learning.

adarsh2209 · Answer

I like the solution - it’s helpful.

Goldenfuture · Answer

Go with options - that's the best approach here I guess

luisdicampo · Answer

Deconstructing the Question  
John needs at least 12 distinct codes.
With  n  colors, he can create:
 n  single-color codes
and
 \binom{n}{2}=\frac{n(n-1)}{2}  two-color codes.

Total codes:
 n+\frac{n(n-1)}{2}=\frac{n^2+n}{2}

Step-by-step  
We need:
 \frac{n^2+n}{2}\ge 12

Test small values:

n=4 :
 \frac{16+4}{2}=10  (not enough)

n=5 :
 \frac{25+5}{2}=15  (enough)

Minimum number of colors is:

Answer: 5

M04-29

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