Math Q: Combination Formula??

[Sprite_TM]

hi there,

the question is from kaplan 2009 premier Q50 in Chapter 9

In a certain laboratory, chemicals are identified by a color coding system. there are 20 different chemicals. Each one is coded with either a single color or unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

a)5
b)6
c) 7
d) 20
e) 40

i read the explanation and i don't understand what the combinations formula is and when to use it. can someone explain how to arrive at the solution for me? thanks

[matt7898]

the way in which to solve this is to break it down into two different combinations. You have to find out how many combinations of colors you can get from choosing one or two from each choice.

For example using 6 which would be the right answer you would have to compute 6c1 and 6c2 so it would be 6!/(5!*1!) + 6!/(3!*2!) which equals 21

[awesomeusername]

I don't know what the answer is, but here's my guess:

I believe the combinatorics formula is:

nCr = n!/(n-r)!

meaning "n choose r" combinations. For example, if there are 5 different colored marbles, how many cobinations of pairs (2 marbles) can be chosen? This would be 5C2, or 5 choose 2. Using the formula, you get 5C2 = 5!/(5-2)! = 5!/3! = 1*2*3*4*5/1*2*3 = 4*5 = 20.

Now let's relate this equation to our problem at hand. It states that the order of the color pair does not matter. This means that Blue-Green is the same thing as Green-Blue. Since the formula above does take into account order, you will have to divide by two the answer to remove this.

Also, since the question states that single colors are possible too, we know that the number to denote the single color is "n" in nCr. "n" is precisely the different number of colors that we will be choosing out of. We already know how to get all the two number combincations:

That's n!/(n-r)! DIVIDED by 2 (because we don't care about order). So it's
n!/2(n-r)!. And since we know that n is the total number of single colors, we add n to that expression to get:

n!/2(n-r)! + n.

We know that this will yield at least 20 combinations. So the expression will finally be:

n!/2(n-r)! + n >=20. We know that r=2 because we are choosing 2 colors. So:

n!/2(n-2)! + n >=20.

Looking at the answers you gave, it's obvious that d and e are wrong because you don't even have to use two-color combinations to denote the chemicals.

If we try 5, then the expression will be:
5!/2(3!) + 5 = 15, which is NOT greater than or equal to 20.

Now let's try 6.
6!!/2(4!) + 6 = 21. Bingo, that's the answer because 21 >= 20.

So the answer is 6, B.

I think I may have went off on a tangent...

[Scott@TargetTestPrep]

hi there,

the question is from kaplan 2009 premier Q50 in Chapter 9

In a certain laboratory, chemicals are identified by a color coding system. there are 20 different chemicals. Each one is coded with either a single color or unique two-color pair. If the order of colors in the pairs doesn't matter, what is the minimum number of different colors needed to code all 20 chemicals with either a single color or a unique pair of colors?

a)5
b)6
c) 7
d) 20
e) 40

Let's analyze the answer choices.

If there are 5 different colors, the number of codes that can be formed is 5C1 + 5C2 = 5 + 10 = 15, which is not sufficient for the 20 different chemicals.

If there are 6 different colors, the number of codes that can be formed is 6C1 + 6C2 = 6 + 15 = 21, which IS sufficient for the 20 different chemicals. Therefore, we need at least 6 different colors.

Answer: B