Wikipedia:Reference desk/Archives/Mathematics/2007 August 3

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August 3

Permutations

I was asked how many possibilities there are when given 10 true/false questions. So there are 10 questions, with 2 outcomes for each one. How do I plug this into the nPr equation? It seems logical to use 2P10, but I can't factorial a negative number. What do I do???

Permutations has nothing to do with this problem. Permutations is when you are picking an order for something, but we aren't picking an order, we're picking outcomes? Start small... how many outcomes are there for 2 true-false questions? 3? etc. Then see if you can see the pattern. This isn't homework, right? Gscshoyru 14:50, 3 August 2007 (UTC)

It is more like "permutations with repetition allowed" than like combinations with or without repetition. Numbering the questions from 1 to 10 and taking the answers to be "T" or "F", one of the many possibilities is "1:T 2:T 3:T 4:F 5:T 6:F 7:F 8:F 9:T 10:F", which we can abbreviate to "TTTFTFFFTF" by leaving out the question numbers. This possibility is different from "FTFFFTFTTT", which consists of the same outcomes in a different order. So order does matter here. For a similar question, which can help to find the answer, consider the case of 2 questions with 10 outcomes "0" through "9". The possibilities are "00", "01", "02", ..., "09", "10", "11", "12", ..., "97", "98", "99". So the number of these is X, which is equal to Y to the power Z (for the right X, Y and Z).  --Lambiam 16:05, 3 August 2007 (UTC)

Use the Fundamental Counting Principle or the "FCP". Question 1 has 2 possible choices: true or false. Question 2 has 2 possible choices: true or false. Same for Questions 3, 4, 5, 6, 7, 8, 9, and 10. The FCP states that the total number of possibilities is the multiplication of all of these ten distinct facts. Therefore, there are 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 possible outcomes. Or, more simply stated: 2 to the 10th power. Incidentally: only one of these possibilities is the correct possibility (i.e., think of the correct Answer Key) ... and all of the other possibilities (2 to the tenth power minus 1) are incorrect possibilities (i.e., at least 1 answer to the ten questions -- or perhaps more than one -- is the wrong answer).

If this does not make sense to you, apply the logic that Gscshoyru (above) proposed. Let's say there are only two true/false questions on the exam. The first question has 2 possibilities: true or false. The second question has 2 possibilities: true or false. So, there are 2 x 2 = 4 total possibilities. Namely, (1) Question 1 is True and Question 2 is also True; (2) Question 1 is False and Question 2 is also False; (3) Question 1 is True and Question 2 is False; or (4) Question 1 is False and Question 2 is True. These are the only 4 (2x2) possibilities that a student can use as answers to the exam. In shorthand, TT or FF or TF or FT. And, one of those four possibilities is the correct answer key (that test would score 100%) and the other three of those four possibilities has at least one incorrect answer found within it (that test would NOT score 100%). So, as this logic applies to a 2-question exam ... it can be applied to a 3-question exam, a 100-question exam, or -- in your case -- a 10-question exam. (Joseph A. Spadaro 02:13, 4 August 2007 (UTC))

See Exponentiation#Combinatorial_interpretation. Bo Jacoby 14:13, 4 August 2007 (UTC).

Does commutativity imply associativity?

I thought it did, but I can't think of an obvious proof. Does it only hold for certain kinds of operations? If not, what's an example of an operation that's commutative but not associative? —Keenan Pepper 19:11, 3 August 2007 (UTC)

One does not imply the other, they are separate. See Example_of_a_commutative_non-associative_magma Gscshoyru 19:21, 3 August 2007 (UTC)

An important example is floating point addition. Although x +_fl y does equal y +_fl x, the sum of two small numbers plus a large one can be greater than the sum of the large one plus each small one in turn. For example, in IEEE 754 arithmetic let x = 1.0 and let y = 1.0e−16; then (x +_fl y) +_fl y equals 1.0 (the same as x), but x +_fl (y +_fl y) is slightly greater. This has practical implications in numerical software; see Kahan summation algorithm. --KSmrq^T 21:00, 3 August 2007 (UTC)

No. Otherwise there would not be two different terms.--SpectrumAnalyser 21:17, 3 August 2007 (UTC)

That is a non sequitur, since there might be (and are) operations that are associative but not commutative. For two properties to be equivalent, the implications have to go both ways.  --Lambiam 21:25, 3 August 2007 (UTC)

Yes I know. I was just trying to keep it simple. (KISS)--SpectrumAnalyser 21:36, 3 August 2007 (UTC)

"Everything should be made as simple as possible, but no simpler." —attributed to Albert Einstein —Keenan Pepper 02:24, 5 August 2007 (UTC)

I think the simplest example is neither RPS nor floating-point addition, but rather just the average (the binary operation which returns the arithmetic mean of its two operands). -- Meni Rosenfeld (talk) 12:28, 5 August 2007 (UTC)