A company that ships boxes to a total of 12 distribution centers uses

Similar questions to practice:
https://gmatclub.com/forum/each-student ... 51945.html
https://gmatclub.com/forum/all-of-the-s ... 26630.html
https://gmatclub.com/forum/if-a-code-wo ... 26652.html
https://gmatclub.com/forum/a-4-letter-c ... 59065.html
https://gmatclub.com/forum/a-5-digit-co ... 32263.html
https://gmatclub.com/forum/a-company-th ... 95946.html
https://gmatclub.com/forum/a-company-pl ... 69248.html
https://gmatclub.com/forum/the-security ... 09932.html
https://gmatclub.com/forum/all-of-the-b ... 50820.html
https://gmatclub.com/forum/a-local-bank ... 98109.html
https://gmatclub.com/forum/a-researcher ... 34584.html
https://gmatclub.com/forum/baker-s-doze ... l#p1057502
https://gmatclub.com/forum/in-a-certain ... 36646.html
https://gmatclub.com/forum/m04q29-color ... 70074.html
https://gmatclub.com/forum/john-has-12- ... 07307.html
https://gmatclub.com/forum/how-many-4-d ... 01874.html
https://gmatclub.com/forum/a-certain-st ... 85831.html
https://gmatclub.com/forum/the-simplast ... 05845.html
https://gmatclub.com/forum/m04q29-color ... 70074.html
https://gmatclub.com/forum/the-telephon ... 20252.html

Let the colors be n. We need to find the min value of n for which nC2 >= 12.

Using answer choices, if n = 5; 5C2 = 10 so we are short by 2.
If n = 6, 6C2 = 15. Good!

Answer (C).

The statement says a single color or a pair of two different colors is chosen to represent each center.
Hence, the number of combinations have to be greater or equal than 12:

\(nC1 + nC2 >= 12\)

Where:

\(nC1 = n\)

\(nC2 = \frac{n*(n-1)}{2}\)

So:

\(nC1 + nC2 = n+\frac{n*(n-1)}{2} >= 12\)

\(\frac{2n+n*(n-1)}{2} >= 12\)

\(2n+n*(n-1) >= 24\)

\(n2+n >= 24\)

We can now pick values for n, which will be faster than solving:

If n=4,

\(n2+n = 20 < 24\)

If n=5,

\(n2+n = 29 >= 24\)

Answer: B (n=5)

For those willing to solve for n:

\(n2+n >= 24\)

\(n2+n -24 >0\)

Solving for n,

\(\frac{-1+-sqroot(1+96)}{2}\)

\(\frac{-1+-sqroot(97)}{2}\)


We don't really need to solve the square root:

- Negative option of the square root is not possible
- The positive option can be approximated by

\(\frac{1+sqroot(100)}{2} = (approx.)= 5\)

(but slightly less than 5)

Given that the inequality is:

\(n2+n -24 >0\)

Any value of n>=(slightly less than 5) will make the inequality positive.

Hence n=5

Answer: B (n=5)

Can anyone tell me when we use n^2 and when we use n+nC2 ????

In this color question we use n+nC2 >= 12

In integer questions we use n^2>=15 ....

Hi gmathopeful90,

The "restrictions" in the question are what dictate the math.

Consider these possible scenarios:

1) You have 5 different colors to choose from and two different rooms to paint. You can use the same color in both rooms. How many different color combinations are there for the two rooms?

Here, the first room could be 5 different colors and the second room could be 5 different colors, so (5)(5) = 5^2 = 25 options.

2) You have 5 different colors to choose from and two different rooms to paint. You CANNOT use the same color in both rooms. How many different color combinations are there for the two rooms?

Here, the first room could be 5 different colors; once you assign that first color, the second room could only be 4 different colors, so (5)(4) = 20 options.

3) You have 5 different colors to choose from. How many different 1-color and 2-color codes can you form with the following restrictions: the 2-color codes must use 2 DIFFERENT colors and the order of the colors does not matter (so blue-green is the SAME code as green-blue)?

Here, you start with the 5 different 1-color codes, then 5c2 different 2-color codes = 5 + 10 = 15 codes.

GMAT assassins aren't born, they're made,
Rich

Thanks for the reply

Do you mean in questions where we assume order of colors in cominations matters, we can use 5*4..
But where color doesn't matter, we use 5C2 ???

This explains stuff for me

Hi gmathopeful90,

You've hit on THE key difference between Permutation and Combination questions: does the order MATTER or not.

IF you're putting things in order (the word "arrange" or "arrangements" often shows up in these types of questions), then you have to keep track of the number of options at each "step" and standard multiplication is involved.

IF you're picking combinations of things (the word "combination" is the common word in these questions), then the order of the items does NOT matter and you have to use the Combination Formula.

One of the interesting "design elements" of Official GMAT questions is that you can use either of the above approaches on certain types of prompts - you just have to be careful about how you set up the math (and you have to be really organized with your work).

GMAT assassins aren't born, they're made,
Rich

I labeled the colors alphabetically and wrote them out

A
B
AB
C
CA,CB,
D
DA,DB,DC 4 Colors (ABCD) = 10 combinations so 1 more color will give more than 12 combinations.

Hi DJ1986,

The 'brute force' approach that you used is PERFECT for these types of questions. When the answer choices are relatively small, it can sometimes be fastest/easiest to just put pen-to-pad and 'map out' all of the possibilities. In that way, you're not trying to make the solution overly-complicated and you're not starting at the screen (hoping that some idea will come to you). You'll likely find that you can take this approach on a few questions in the Quant section on Test Day, so don't be shy about using it (and practicing with it in mind).

GMAT assassins aren't born, they're made,
Rich

I am confused with the line 'order doesn't matter'.

Does it mean BR = RB or BR =/ to RB?.

Hi ameyaprabhu,

When the order doesn't matter, RB and BR are the SAME option (so you can't count it twice, you can only count it once). In these sorts of questions, it can often be fastest to just 'list out' the possibilities (as opposed to doing lots of complex calculations).

GMAT assassins aren't born, they're made,
Rich

Could you please explain me how you get [fraction]n(n-1)/2 from C^2_n? Shouldn't it be [fraction]n!/k!(n-k)! ?
Thanks

\(C^2_n=\frac{n!}{(n-2)!*2!}=\frac{(n-2)!*(n-1)*n}{(n-2)!*2!}=\frac{(n-1)*n}{2}\).

Hope it's clear.

1. Solving a simple case and then generalizing would be easy for this problem.
2. Take 2 colors Red and Blue. These two can be used in the following ways R, B, RB. i.e, 2+2C2. It can represent only 3 centers
3. Take 3 colors R, B, G. These can represent 3 +3c2=6 centers
4. Four colors can represent 4+4C2= 10 centers
5 colors can represent 5+5C2=15 centers

So we see a minimum of 5 colors are needed

Since we have only 12 distribution centers, we know we will need fewer than 12 different colors to identify them.

Let’s say we have 4 different colors; then 4C1 = 4 centers can be identified by one color, and 4C2 = 6 centers can be identified by two different colors. So a total of 4 + 6 = 10 centers can be identified.

We see that if we have only 4 different colors, we don’t have enough ID codes to assign to the 12 centers. Therefore, we need one more color.

If we have 5 different colors, then 5C1 = 5 centers can be identified by one color, and 5C2 = 10 centers can be identified by two different colors. So a total of 5 + 10 = 15 centers can be identified.

We see that if we have 5 different colors, we have more than enough ID codes to assign to the 12 centers.

Answer: B

The first order of business is to try to determine whether this is an exponential, permutation, or combination, since that will affect the equation you use. Since the question states explicitly that order does not matter, you know you will be applying the combinations formula.

C=N!/K!∗(N−K)!
Reverse-engineering questions generally require you to plug in answer choices to find the one that works. You need to find an answer that sums to at least 12 arrangements, since you need to code 12 separate distribution centers. You need to do a calculation for one letter (which does not really need a formula!) and for two letters as shown below - when the smallest value reaches at least 12 you have the right answer. Let’s start with “B”. Plugging in 5 for N, we see that:

A=5!/1!∗(5−1)! + 5!/2!∗(5−2)! = 5+10 = 15

With N = 5, you have more than enough for the 12 distribution centers, but only just barely; it seems unlikely that dropping down another number would still keep it above the minimum. However, just in case, here is what it would look like if N = 4 (answer choice “A”):

A=4!/1!∗(4−1)! + 4!/2!∗(4−2)! = 4+6 = 10

Answer choice “A” drops the value below the minimum possible (we need codes for 12 distribution centers). Therefore, the minimum value for N is N = 5, and the answer is “B”.

For those like gmathopeful90 who may need formula usage criterion for n^2, I found this:

Ordered?-----Repetition? --- Formula

Yes (permutation)----No---------P(n,r)=\(\frac{n!}{(n−r)!}\)

No (combination)----No---------C(n,r)=\(\frac{n!}{r!(n−r)!}\)

Yes (permutation)----Yes--------P(n,r)\(n^r\)

No (combination)----Yes--------C(n+r−1,r)=\(\frac{(n+r−1)!}{r!(n−1)!}\)