Choosing 3 of 6 ColorsDate: 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/ |
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