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### Choosing 3 of 6 Colors

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Date: 03/03/2000 at 00:33:35
From: Cheyenne Marcy
Subject: Different ways to pick 3 crayons out of 6

My math problem of the week is this:

Miss Davis asked her students each to draw a picture describing their
hobby. She asked them to use only three colors. Patrick had a box of
crayons containing the colors red, blue, yellow, orange, green and
purple. How many different ways can Patrick use his crayons to draw
his picture? The hint is to use an organized list.

I listed the colors three at a time like this:

rby, rbo, rbg, rbp, byo, byg, byp, byr, yog, yop, yor, yob,
ogp, ogr, ogb, ogy, gpr, gpb, gpy, gpo, prb, pry, pro, prg.

Then I went through the list and crossed out duplicates and I got:

rby, rbo, rbg, rbp, byo, byg, byp, yog, yop, yor, ogp, ogr,
ogb, gpr, gpb, gpy, pry, pro.

My mom helped me, but we're not sure we got them all. She says there
is a formula for combinations or permutations that she learned in
college, but she is not sure which formula would work and she doesn't
remember the formulas anyway.

Is my way right?
```

```
Date: 03/03/2000 at 07:06:17
From: Doctor Anthony
Subject: Re: Different ways to pick 3 crayons out of 6

The answer is 20 so you found most of them. The formula your mother
was trying to remember is C(6,3) the number of ways that 3 things can
be chosen from 6 different things.

6 x 5 x 4
The formula for calculating C(6,3) is   ------------  =  20
1 x 2 x 3

This formula is derived as follows:

You can choose the first color in 6 ways from the 6 available. For the
second choice you now have 5 colors to choose from, so there are 5
ways to choose the second color and finally 4 ways to choose the third
color. So the top line of our calculation is

6 x 5 x 4 = 120

We must now show why this answer is too large by a factor of 1 x 2 x
3.

Suppose one of our choices was rby. Then other choices could be ybr,
or byr or ryb and so on. In fact with 3 different colors there are 6
different ways of arranging them. (We could choose first position in 3
ways, second position in 2 ways, third position in 1 way giving
3 x 2 x 1 = 6 ways). But ALL 6 of these arrangements still count as
ONE choice of 3 colors, and so the number 120 is too large by a factor
of 6.

It follows that the number of groups of 3 colors that can be chosen
from 6 different colors is:

6 x 5 x 4       120
-----------  =  -----  =  20
1 x 2 x 3        6

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 03/03/2000 at 15:45:14
From: PJ Marcy
Subject: Re: Different ways to pick 3 crayons out of 6

Okay, I understand your explanation for this particular problem, but
I'm not sure how to explain to my daughter (1) when this formula is
applicable, and (2) how to arrive at the numbers to use.

"C(6,3)" is not representative of anything to me. It would seem that
from C(6,3) you are deriving (C-N)!/N!, where C is the total number of
crayons and N is the number of crayons in use. Is this correct? If so,
I think that I can say to my daughter, "Any time you have a set of
available choices and a subset of allowable choices, ..."

Also, this is stuff I had in a college statistics class. My daughter
is in 4th grade! My daughter is pretty exceptional in math. If I can
explain it, or her teacher can, she can probably grasp it (at least
for the moment). But is this something she and other 4th graders
should be trying to "wrap their brains around?"

Thanks for your help.
P. J. Marcy, Cheyenne's mom
```

```
Date: 03/03/2000 at 18:04:15
From: Doctor Anthony
Subject: Re: Different ways to pick 3 crayons out of 6

The formula for C(6,3) is usually stated in the form

6!        6 x 5 x 4 x 3!       6 x 5 x 4
-------  =  ---------------  =  ------------  = 20
3! 3!       1 x 2 x 3 x 3!       1 x 2 x 3

n!
And the general formula is  C(n,r) =  -------
r!(n-r)!

However, clearly it would not be appropriate to quote the expression
in this form to a 4th grader.

C(n,r) is the number of 'combinations' of r things that can be chosen
from n different things. It is used when the order within the
combination is not important. For example choose 4 people from 10 to
be passengers in one car, while the other 6 go in a minibus. The
number of ways of choosing the 4 people is C(10,4) = 210

If the order does matter then we use P(n,r). In this case the
calculation of P(6,3) is simply  6 x 5 x 4  (start at 6 and go for 3
factors). An example might be the number of ways of selecting a
chairman, vice-chairman and secretary from a group of 6 people. In
this case the order does matter because the jobs on the committee are
different.

How much of this you impart to your daughter is something only you can
decide. Combinatorics requires VERY clear thinking rather than
advanced training in mathematical techniques, so it is an appropriate
topic for bright children as a mental exercise.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Elementary Puzzles
Elementary Word Problems
High School Permutations and Combinations
High School Puzzles
Middle School Factorials
Middle School Puzzles
Middle School Word Problems

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