At a certain laboratory, chemical substance are identified by an unord

Question

At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using seven colors?

A) 21
B) 35
C) 105
D) 135
E) 210

Looking for easier method to solve this.

A) 21
B) 35
C) 105
D) 135
E) 210

Looking for easier method to solve this.

BrushMyQuant · Answer

Since no chemical may be assigned the same three colors as any other. So order does matter. So answer should be 7C3 = 35 ?

EMPOWERgmatRichC · Answer

Hi amidamani13,

You mentioned in your post that you were looking for an "easier" method to solve this problem. What method were using before (or what method did the explanation that came with this question use?)?

"Combination" questions are a rarer category on Test Day (you'll see 1, but you might see 1-2 more). In this prompt, the words "unordered" and "combinations" provide the clues to let you know that you can use the Combination Formula to answer the question.

Combinations = N!/  where N is the total number of items and K is the size of the subgroup

Here, we have 7 colors and we're choosing combinations of 3. Plugging those values in, we get...

7!/(3!4!) = (7)(6)(5)/(3)(2)(1) = (7)(5) = 35 combinations of 3 colors.

Final Answer:  B

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

Bunuel · Answer

Check  Constructing Numbers, Codes and Passwords  in our  Special Questions Directory .

rsh12 · Answer

Hello Bunuel,

Since the question says it is unordered and no chemical can have same 3 colors -> This should not omit the possibility of 2 same colors in 3 color combination right?
I was looking for an option which has 7*7*5 maximum number of substances.
Isn't there something wrong in the question unless I am not understanding it correctly?

chetan2u · Answer

Hi rachit,
7*7*5 as per you also includes combination of two same colour and one different colour also.
But look at the coloured portion, it clearly specifies three different colour...
your solution would be close to correct if it said that all three colours can not be the same

sidoknowia · Answer

3 out of 7 can be chosen in 7C3 ways
chosen 3 colors can be arranged in 3! ways

so answer should be 7c3 * 3!

Am I correct? if not, what am I doing wrong?

EMPOWERgmatRichC · Answer

Hi sidoknowia,

The prompt tells us that substance is identified by an UNORDERED combination of 3 colors and that no chemical may be assigned the same 3 colors as any other chemical.

As an example, a chemical with the colors A/B/C means that the following combinations CANNOT be used by any other chemical:

ABC
ACB
BAC
BCA
CAB
CBA

By extension, the first part of your calculation would be the number of possible substances that could be identified with 7 colors: 7c3 = 35 possible substances

Final Answer:  B

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

MathRevolution · Answer

The total colors are 7 and to be chosen are 3.

=>  ^{7}{C_3}

=> 7 * 5 = 35

Answer B

prateekchugh · Answer

At a certain laboratory, chemical substance are identified by an unordered combination of three different colors. If no chemical may be assigned the same three colors as any other, what is the maximum number of substances that can be identified using seven colors?

A) 21
B) 35
C) 105
D) 135
E) 210

Looking for easier method to solve this.

Unordered combination: Formula---nCr=  \frac{n!}{r!*(n-r)!}

We have total 7 colors and need to choose 3 colors= 7C3=  \frac{7!}{3!*4!} =35

A) 21
 B) 35 
C) 105
D) 135
E) 210

Fdambro294 · Answer

I started off making it way too complicated:

I misread the Q-Stem and though that we could have a Code where the 3 Colors are all Different and a Code where 2 are the Same (with the 3rd Different).

However, the Q says: "unordered combination of 3 DIFFERENT colors" for Each Code.

The Maximum amount of Codes we can make is simply the No. of Different Ways we can Form a Group of 3 Different Colors out of a Total of 7 Colors --> to make 1 Code. This is the Combination Formula.

"7-choose-3" = 7! / (3! * 4!) = 35 Maximum Codes

-B-

ScottTargetTestPrep · Answer

Solution:
 
The number of ways of choosing 3 colors out of 7 (where order doesn’t matter) is 7C3 = 7! / (3! x 4!) = (7 x 6 x 5) / (3 x 2) = 7 x 5 = 35.

Answer: B

Nikhil30 · Answer

it it wrong because what you have written is same as writing npr. secondly here its written unordered manner hence order doesnt matter here so ncr. third for example: if a color code of a chemical is ABC than there should not be any color code as ABC,CBA,BCA,ACB.......etc etc. thats why you shoule eliminate 3!. hope its clear

Ashish2120 · Answer

What is the meaning of "If no chemical may be assigned the same three colors as any other" in the q.

Posted from my mobile device

EMPOWERgmatRichC · Answer

Hi Ashish2120,

When the prompt states that each substance is identified by an UNORDERED combination of 3 colors and that no chemical may be assigned the same 3 colors as any other chemical, it means that each group must be unique.

For example, a chemical with the colors A/B/C means that the following combinations CANNOT be used by any other chemical:

ABC
ACB
BAC
BCA
CAB
CBA

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

Ashish2120 · Answer

Thanks is it the other way of saying that there is no repetition. I am confused, if it is already written that it is unordered. Why this second line.

Posted from my mobile device

PolarisTestPrep · Answer

Hello,

This question is asking how many ways there are to group seven colors together in groups of three.

How many different combinations of 3 colors can be made from 7 total colors?

This is a combinations problem.

The formula for combinations is   nCr = \frac{n!}{r!(n-r)!}   where n is the total number of objects in the set and r is the number you are going to combine at one time

In this case, n = 7 different colors and r = combinations of 3 colors to identify a chemical

\frac{7!}{3!(7-3)!}

\frac{7!}{3!4!}

\frac{7*6*5*4*3*2*1}{3*2*1(4*3*2*1)}  =  \frac{7*6*5}{3*2*1}  when you cancel (4*3*2*1) from top and bottom

This equals  \frac{210}{6}  =  \frac{70}{2}  = 35

The answer is (B)

EMPOWERgmatRichC · Answer

Hi Ashish2120,

The second line means that NO other chemical can be identified by that exact same combination of 3 colors.

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

achloes · Answer

I think that simplifying the stem is key to getting this question right.

We are told the following: 
- there are 7 colors
- need to make unique "codes" of 3 with the 7 colors 
- the order of colors for the codes do not matter

We're asked how many of such codes can be made aka how many substances can be uniquely identified.

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