How many 3-digit numbers can be formed using only 2, 3, 4, 7 and 8 with no repeats?
There are various ways to tackle this problem (which can easily be generalized for n-digit number or words within a given alphabet).You have 3 places to fill and have 5 digits that you can use but can't repeat. So you have 5 options for the first place, leaving 4 choices to fill the other 2 places, then you choose one of the four and have 3 option for the last place. So you have [math]5 * 4 * 3 = 60[/math] ways to get your desired number.You can also see this from other perspective, you have 5 digits and you will only use (any) 3 of them, that's [math]\binom{5}{3} = 5! / (5!(5 - 3)!) = 5! / (3! * 2!) = 10[/math] of choosing those 3 digits. But you can arrange these 3 digits in any way you want, that is, any permutation of them: [math]3! = 6[/math]. Which, as expected, also gives [math]\binom{5}{3} * 3! = 60[/math] solutions.One other possibility would be to use the inclusion-exclusion principle which, for this example, would be too boring. (If you want, ask in the comments and I'll write how to do it here).Hope this answer helps you!
SAT math problem word problem/ probability question help!!?
The method you used is for permutation problems which is when ordering is important. This is a combination problem since the order in which you select the plumbers does not affect the team assembled Obviously there are four ways to select the experienced plumber. However for the trainee selection consider this - let them be called A, B, C and D. Then from your argument there are four ways to pick the first trainee, say A, then we are left with B, C and D which means there are three ways for the second selection. Say we pick B, then the team assembled would consist of A and B. However, we could have assembled this same team by picking B first and then A which would have been a different permutation but would result in the exact same team. So listing all the possible permutations we would have 12 permutations as follow, AB BA AC CA AD DA BC CB BD DB CD DC Notice that two would consist of the exact same member differing only in their order of selection, so there would not be 12 selections for the trainee but rather half as much which is 6 ways. Using the counting principle we would get the number of possible ways to form a group of one experienced plumber and two trainees equal to 4x6 = 24 ways. In general when doing a permutation involving a selection of r items from a group of n items the number of possible permutations is (n!/r!) The number of possible combinations would be (n!/r!) / (n-r)! where (n-r)! is the number of ways in which you could rearrange the items selected. Using the given problem as an example there are 4 trainees and we want to select 2 of them. The number of permutations is 4! / 2! = (4x3x2x1) / (2x1) = 4x3 = 12 Now when making two selections there are (4-2)! = 2 (or equivalently 2x1) ways in arranging those items so we divide the number of possible rearrangements from the number of possible permutations to obtain the number of possible combinations.
Pre-cal world problem help please!!?
1. You have relatives living in the United Kingdom and in France. Suppose that you have purchased a prepaid phone card with a value of $75. Calls to the United Kingdom cost 23 cents per minute, while calls at France cost 21 cents per minute. Part 1: Write an expression for the total cost of calling to the United Kingdom, U(x), for a total of x minutes. Part 2: Write an expression for the total cost of calling to France, F(y), for a total of y minutes. Part 3: Write a linear inequality involving x and y to represent the number of minutes you can use to call the two countries and not exceed the value of your prepaid calling card. Please show all your work :) Thank you for your help!
Algebraic solutions to 4-digit numbers of the form: a^b*c^d = abcd?
Point 1; i haven't worked with mods a lot, but i guess I can still add to the ideas here. :D So we have a^b*c^d = 1000a + 100b + 10 c + d. since abcd is divisible by a, If we subtract the number by 1000a, this number will still be divisible by a. Similarly, if we subtract by 10c, then 1000a + 100b + d is divisible by c. Now, if you can prove that b or d cannot = 1, then you can take this further because it would be at least divisible by a^2 or c^2. I'm not sure if eliminating 1 variable would help. Point 2: Somehow 'simplify' this equation to (ac)^b * c^(b-d) = abcd. Not sure if that helps either. Sorry to disappoint if I got you all excited, but I just thought about this and I'll add to it when I can. (And I'll start doing some reading on Mods) Edit 1: All numbers squared would give an ending digit of either 0,1,4,5,6 or 9. Now looking at this: 0 and 5 can't work as you posted this(and I can see that it's not hard to prove) So we are left with numbers ending in 1 4 6 or 9. Assuming that b and d are both even, we can determine that the 4 digit number is a perfect square. Meaning that d = 4 or 6. Now I'm thinking of a way to prove that b and d cannot both be even. (Sorry if my working is all over the place - I am trying to bring up new ideas) Edit 2: Or is there a way to prove that you can't prove it? I know it sounds a little hard for this type of question though. Edit: Yeah all those bits are assuming that b and d are both even - in which case the number will be a perfect square. If we can counter prove that then we get an odd c and an even d. Edit again: If we are given that a=d, will that help you? If you can solve a^b * c^a = abca, that can possibly lead to solving this problem. Possibly last edit before a breakthrough: I think that we need to compromise and take a few steps backwards. Lets change this question to a^b * c^a = abca, where 2a +b = c, which makes it a^b * (2a+b)^a = abca, where abca = 1000a + 100b + 10(2a+b) +a = 1021a + 110b. If we can't solve a^b * (2a+b)^a= 1021a + 110b, then we're pretty much done for. (Note all these changes can apply to the number 2592.) Edit: Oversimplifying is the whole point. If we can't even prove the simplified problem, there is no way we can prove the original.
What are some examples of problems which are: (1) NP but not NP-Complete; (2) NP-Complete; (3) NP-Hard but not NP-Complete?
You first have to understand the meaning of NP, NP-Hard and NP-Complete.Giving definitions straight from Wikipedia:NP is a "class of computational problems for which solutions can be computed by a non-deterministic Turing machine in polynomial time. Or, equivalently, those problems for which solutions can be checked in polynomial time by a deterministic Turing machine."Another way to say this is that given a solution to this problem, it can be verified in polynomial time.NP-Hard problems are a "class of problems which are at least as hard as the hardest problems in NP. Problems in NP-hard do not have to be elements of NP, indeed, they may not even be decision problems."NP-Complete problems are a "class of problems which contains the hardest problems in NP. Each element of NP-complete has to be an element of NP."Another way to say this is that a problem is NP-Complete if it is NP-Hard and given a solution to any instance of the problem, the solution can be verified in polynomial time.So an example of a problem in NP but not NP-Complete is the sorting problem. i.e. Given [math]n[/math] integers, rearrange the numbers such that they are in non-decreasing order. This can be easily solved in [math]O(n \log{n})[/math] (well, actually better). Clearly, you can verify if a proposed solution is actually a solution in [math]O(n)[/math], which is polynomial in [math]n[/math].NO problem can be NP-Complete, without being NP-Hard by the definition of NP-Completeness.An example of an NP-Complete problem is clique. i.e. Given an undirected graph, what is the largest complete graph which is a subgraph of the graph.Now, for an NP-Hard problem that is not in NP (i.e. not NP-Complete). Given some [math]n[/math], find all cliques in all graphs with [math]n[/math] vertices. Clearly this problem is harder than the previous clique problem since if we can solve this problem, we can solve the previous clique problem. Also, note that the answer to this problem is actually all subsets of the [math]n[/math] vertices which form cliques. But also note that to verify that we have the correct answer, we have to check that we have all subsets which form cliques. In fact, this problem is NP-Hard, but not NP-Complete.See: Np Complete
How Do I get 2,100 Minimum on the SAT?
I've been studying a LOT this month. I have summer vacation until September 8th. Basically, I do a practice SAT 5 days a week (I've done 23 so far), 8 from the college board book I've done, 11 from the 11 Practice SAT Princeton book, and 4 more from the Princeton SAT prep course, which I'm almost done with. I revise my mistakes on all the practice exams. I also do about 10 vocab words a day which I make flashcards for out of the Princeton Review book, the most common words. On the College Board ones I got around 1900's. On the Princeton ones I got around 1700's-1800's. I NEED a 2,100 minimum. I need to get into a good medical school and really want to go to any Ivy League school, even the lower-end ones. Basically, my worst area is critical reading. My last Princeton score was like this: Writing: 680 Math: 620 Reading: 500 I hate critical reading. I've never been one who reads a lot or one who reads good literature, except that from school (Lord of the Flies, Beowulf). I can't read quickly on the SAT, sometimes I don't understand part of the passage, or the vocab, or the questions are hard, and all the choices seem good. Math and writing I think I can improve. How do I get a minimum of 2,100??