Question about Fundalmental Counting principle?
405265
there are 17,576,000 possible arrangements. First, take the three letters, and find out all the possible combinations that there are for them. This answer turns out to be 17,576. Then take the numbers. There are 1,000 possible arrangements, so the number of possible arrangements for the license plate is 17,576,000.
352393
There's an early chapter on the continuum hypothesis in Leonard Wapner's "The Pea and the Sun," which is about the Banach-Tarski Theorem.  The wonderful thing about that book is that by the end I actually understood the theorem.\n\nAlso, when I was quite young I read Lilian Lieber's book "Infinity."  She was a mathematician and took it far beyond a high school student's level, yet it was a fascinating read.  Understood why Cantor went nuts.
628612
10 - Some people just can't count the way you can!\n\n(I'm looking forward to commenting on this question, assuming other people know what you're even talking about!!!!)
294686
The number of permutations of a string of n unique objects is n!. \nThis can be derived by the fact that if you have the set of all permutations of that string, all the elements of the set can be generated by first choosing an object to be first in the string, then a second object from the n-1 remaining objects to be second in the string, and so forth. By the fundamental principle of counting, one can thus generate n*(n-1)*(n-2)*...*1 = n! elements.\nIf some of the objects in the string are not unique, ie., the number of permutations of the string (E, E, E, V, R, S, T), we can show that the number of unique permutations is the same as the number of permutations given by pretending the objects are unique divided by the number of permutations of the non-unique objects.\nThis can be shown as follows: Suppose P(n) is the number of permutations of n unique objects. Suppose that P(n, r) is the number of permutations of n unique objects and r non-unique objects. Then by the fundamental principle of counting, if we replaced each non-unique object in the string from a set of r objects, we have P(n, r)*P(r) = P(n) permutations. Thus P(n, r) = P(n)/P(r) = n!/r!.
339026
my claim is that the number is infinite.  here is the reason, it depends on who you classify as a mathematician and what date you want to start at.  Also i claim that you can move forward in time, which i will claim is infinite, and find more mathematicians bore in march.
595633
When you are dividing by something you are trying to find out how many of a certain amount (say, 3) is in another amount (say, 12). The result for my example, is of course 4, since 3 goes into the number 12 four times.\n\nNow the number 0 is nothing, literally. So if you're dividing into, say 12, you're wondering how much nothing is in 12. That's just not possible, because 12 is all "something".\n\nI hope that's clear enough...I tried.\n\nEdit: The guy above me is wrong.
186035
1) What is the difference between F(n) and f(n)? \n\n2) I cannot comprehend the second condition.\n\nEdit your question, please.
797253
12,000,198 microscopical creatures\n\nit's 7 x 7 x 7 x 7 x 17 x 7 x 6 x 7 = 12,000,198\n\nBy the way, why is this in Words and Wordplay not Math?\nAm I missing something?
903275
The principle of inclusion-exclusion (PIE) is to overcount what you want to count and then subtracting what was overcounted. The advantage of doing this is that it might be easier to do the overcounting and to count what was overcounted, than to directly count what you need to count. Here is a question that illustrates the cleverness of this principle.\n\nSuppose you are given the six letters A, B, C, D, E, and F. How many 3 letter "words" can you make using these letters, repetition allowed, that have at least one E?\nThe brute force method would involve drawing some sort of tree diagram or complicated case work since no simple multiplication will take into account the possible place of the E in the word and the number of E's in the word.\nThe solution applying PIE is much simpler. It is easy to count the total number of words you can make, with no restriction. It is simply 6^3=216, since there are six choices for the first letter, six for the second, and six for the third. However, you overcounted by exactly the words that have no E's, so subtract the number of words without E's. This is also easy to calculate; since you only have 5 choices for the first, second and third letters, there are 5^3=125 words that do not have E. Then the number of words that have at least one E is 216-125=91.\n\nA derangement is a permutation that leaves no element fixed. An example: If your ordered 4-tuple is (1,2,3,4), one derangement would be (2,3,4,1).\nSo the question is how many derangements of a set with 4 elements are there? We will use PIE.\nFirst, we ask, "How many permutations of 4 elements are there?", since the derangements are permutations with a special restriction. This answer is easy; it's 4!=4 x 3 x 2 x 1 = 24, since there are 4 choices for the first element, 3 for the second, 2 for the third and 1 for the last.\nBy how much did we overcount? We counted permutations that fixes elements; one such permutation is (1,4,3,2), since the position of 1 does not change. To count this number, we apply PIE.\nThere are several cases. Suppose we want to count the permutations that fix 1. Since 1 is fixed, there are 3! permutations of the remaining 3 elements. The same is true if we were to fix 2 or 3 or 4. So we subtract 4 x 3!=24.\nHowever, you probably noticed that some permutations were subtracted twice, specifically the permutations that fix more than one element. So we need to apply PIE again to count the number of permutations that fix two elements.\nPick two elements to fix. There are 4C2 ways to do this. Then there are 2! permutations of the remaining elements. We add (4C2)(2!)=12.\nThe next step is to subtract the permutations that were overcounted again. These are the permutations that leave three elements fixed. There are (4C3)(1!)= 4 of these.\nThe last step is to add back the permutations that were overcounted yet again. These permutations leave 4 elements fixed, so there is only (4C4)(0!)=1.\nThe final answer is 4! - 4 x 3! + (4C2)(2!) - (4C3)(1!) + 1 = 9.\nThis can be generalized to n-tuples: n! - (nC1)((n-1)!) + (nC2)((n-2)!) - (nC3)((n-3)!)+ ..., which through some algebraic manipulation becomes the formula you see on MathWorld.
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Hint: If an analytic function is 0 on a set with its accumulation point, then the function is identically 0.\n\nNow, either f or g has to have this property since at each point one or the other is 0.
380941
The theorem to use is the pigeon hole principle.\nOrder all the people in the US by the number of hairs they have. This number will go from 0 to 200000. Next number all people starting from the first one in line, starting from zero. We can assume this guy to be bald, i.e. zero hairs.\n\nNow walk down the line. Whenever you find a person having less hairs than his ranknumber, you have found two persons with the same number of hairs.
12883
how do you count to infinity?
742071
You put tallies in the table.(Small lines in set of five).eg If digit 3 comes 8 times in a distribution, you put four tallies and cross them with fifth tally diagonally and than after a space add three more tallies,because its easy to count set of five in large data.if 8 comes 4 times than you put four tallies.(////)
214242
notice that if a remainder occurs twice, then it will immediately repeat.  hence the period is at most 111.  Yes, the pattern is distinctive, thats because 111 = 1 + 10 + 100\ntry for example base 7 with 1/(111) where 111 is in base 7.
306636
For the nth level, the quantum numbers left are l,m, and s.\nWe have that l goes from 0 to n-1 and is an integer, and m goes from -l to l and is an integer, while s takes on two values +-(1/2). Pauli's exclusion principle says that no two electrons can have exactly the same quantum numbers. So there is one electron for each valid choice of l,m, and s. Now count the possibilities.
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well the number to the utmost left gets added by 11 and in every row 1 gets added more than the previous row so if u see the difference between any two consecutive row is always 12 !!\nthts wat i thnk
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For n = 5? Seriously?\n\nIt is not clear what you are asking.\n\nSuppose that:\n\n1. Your model is of a sequence of independent Bernoulli trials each of which succeeds with probability p.\n\n2. Your null hypothesis H0 is p = 0.3\n\n3. Your alternative hypothesis H1 is p is not 0.3.\n\n4. You plan to observe 5 trials with the intent of accepting or rejecting the null hypothesis with 95% confidence.\n\nThen from p + q = 1 (i.e., q = 1 - p) and 1^5 = (p+q)^5 = p^5 + 5p^4q + 10p^3q^2 + 10p^2q^3 + 5pq^4 + q^5 you know that the probabilities are\n\n5 successes in 5 trials: p^5 = 0.3^5 = 0.00243\n4 succeses in 5 trials: 0.02835\n3 successes: 0.13230\n2 successes: 0.30870\n1 success: 0.36015\n0 successes: 0.16807\n\ni.e., 5 or 4 successes occur <= 5% of the time so you accept H0 on 3 or fewer successes and reject H0 on 4 or 5 successes.\n\nAnother interpretation of your question:\n\nSuppose your model is of independent identical Bernoulli trials each with unknown probability p of success and you would like to estimate p from observing the outcomes of 5 independent trials.  Given k out of 5 successes as the observed outcome for which p can we say that "k" or "k or fewer" or "k or more" successes out of 5 has probability <= 5% of being observed (that is three different questions by the way).  For each of k = 0, 1, ..., 5 you plot the probability of that outcome given p and exclude those p for which the plot drops to 5% or less.