I've tried looking through the textbook and using other resources!! Can you explain how to do the following problems, please? You don't have to provide all the answers, but if you do, can you still teach me (you don't have to)? \n1. How many ways can a three member committee be chosen from a group of seven people?\n2. How many ways can a chairman, vice-chairman, and secretary be chosen from a group of seven people?\n3. How many ways can a student stack textbooks from his six classes in his locker?\n4. How many ways can you choose five starters from a basketball team consisting of ten players?\n5. How many different two-scoop ice cream cones can be made at a store that has six flavors of ice cream?
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to find this I like to almost make a visual. \n1. you have 3 available slots (make 3 dashes) for the 1st dash you have 7 people you can pick from. so place a 7 on dash one. on the second dash, you have already placed 1 person in slot 1, so you would put 6, and finally for the 3rd dash, you would put the nummber 5 because that's the total selection. now multiply these nummbers together. this is called a combination\n2. do the same thing as in nummber 1...3 dashes for the 3 chairs...1st dash you put 7, 2nd 6 and 3rd 5 and multiply them together.\n3. you have 6 slots so 6 dashes slot 1 there's 6 books to choose from, slot 2 there's 5 to choose from, slot 3 there's 4 to choose from, slot 4 there's 3, slot 5 there's 2 and slot 6 there's 1...multiply them nummbers together.\n4.5 slots...10 times 9 times 8 times 7 times 6 \n5.2 slots 1st slot there can be 6 flavors second there can also be 6 however you have to dicide by 2 because chocolate vanilla is the same as a vanilla chocolate
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no, you use nPr i belive, in any case you take 9x8x7x6 because the first one can be any of the nine players, the second can only be 8, since one has already been chosen, the thrid can only be one of the 7 players left, and finally the last can be any of the 6 players left. \n\nin any case the answer is 3024 combinations
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click on the below link, and choose any one of the results. I would advise reading at least 3 of them to get a feel for theory before choosing one of the walkthroughs.\n\nBest of luck to you!
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(a) There are 10 ways to select one couple.  After you select a couple, there are 18 people left to choose from.  So their are 10*18=180 ways.\n\n(b) Pick the first person from any one of 20 people.  To select the second person, you can pick from 18 others (because you can't select the first person's partner).  To pick the third person, you can pick from 16 others (because you can't select the first person or the second person's partner).\n20*18*16=5760.  Since we are selecting a committee, the order does not matter (it doesn't matter who is the "first", "second", and "third").  Since there are 6 ways of ordering a 3-person committee, there are 5760/6=960 ways.
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D)6\n\nNote the pattern:\nIf f(1) = a, f(2) = b, then\nf(3) = b-a\nf(4) = -a\nf(5) = -b\nf(6) = -b+a\nf(7) = a\nf(8) = b\n\nAt f(7), the pattern starts repeating.  Therefore, the six values are f(1) through f(6) above.\nThis really isn't a permutation/combination problem, but you can often find good answers here on Yahoo Answers!  If you have specific problems with which you have difficulty, why dont you post a few related ones in one question here?  I'm sure I or others should be able to help.\n\nYOUR FOLLOW-UP:\nThis is two small problems in one.\nFirst, find the # of ways to choose the order of the backgrounds.  We have 3 choices for the first one, then 2 choices for the second, then 1 left for the third.  Therefore, there are 3*2*1=6 ways to choose the order of the backgrounds.\n{ (b, e, r), (b, r, e), (e, b, r), (e, r, b), (r, b, e), ( r, e, b) }\nSecond, there would be 3*2*1 = 6 ways to interview the business people.  There would be 2*1 = 2 ways to interview the education people, and just 1 way to interview the recreation person.  So that would make 6*2*1 = 12 ways to interview the people after the background order is set.\n\nThe total number of ways is then 6*12=72 ways to interview all 6.
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Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
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1.Let the no.of boys in the family be x+1\nTherefore,each boy has x brothers(excluding the boy himself)  and x sisters\nSo the number of girls in the family is x\nTherefore each girl has x-1 sisters(excluding herslef,of course) and as assumed x+1 brothers\nBut by the problem,\n(x-1)=(x+1)/2\n+.2x-2=x+1\n2x-x=1+2\n+.x=3\nTherefore there are 3 sosters amd 4 boys in the family and the no. of children is 7.\nso A is correct\n2>Let the no. of children who lost 1 teddybear each be x\nTherefore,the no of children who lost two teddy bears each is 32-x\nAccording to the problem\nx+2(32-x)56\n=>x+64-2x=56 \nor x=8\nTherefore,8 children lost one teddy bear each\nA is correct
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well in my skool we take it as {<and rule>} and {<or rule>example:there is 5 white marble and 5 black marbles in a bag...u pick 2 randomly without replacing them in the bag.....wat is the probability that u get 2 white marbles==>this is a combination 2 the way it is solved is multiplying the the 2 combined probabilities\nanswer:pr:(white AND white)=(5/10)*(4/9)\n\nnow the set is this way....wat is the probability that u get 2 marbles of different colors in the first 2 picks===>set rule\n\nanswer:there is 2 combinations in which u can get 2 different colors there are {white then black} OR {black then white} u find the 2 combinations then ADD them\npr(WandB OR BandW)={(5/10)*(5/9)}+{(5/10)*(5/9)}
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Im assuming that by using the word 'groups', (plural) you are not allowed all four in one group as a solution.\nSo, if the counters are the same, then 2 ways:\n1, 1, 1, 1 \nand\n2, 2\n\nIf the counters are different, then 4 ways.  (If the counters are:\na, b, c, d then:\nab,cd\nac,bd\nad, bc\na, b, c, d. )\n\nIf this is for a 3rd grader then the former is probably what they are looking for, but I can't really know that.  The latter seems a little complicated for the average 3rd grader.
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The Cube has 43,252,003,274,489,856,000 different possible configurations.\n\nI cheated and looked up the answer, but to figure it out you'll want to use combinatorics.  I'll have to work on that part of it...\n\nIf you're taking the cube apart and counting how many different ways you can put it back together, this is a relatively simple problem (although it doesn't match the above answer).\n\nThe center cubes don't move at all, so there is only one possible configuration of them.\n\nThere are 12 edge cubes, which each have 2 outside edges and therefore two orientations for each location.  The number of possible combinations of these edges is (2*12)*(2*11)*...*(2*2)*(2*1) = (2^12)*12!\n\nThere are 8 corner cubes, which each have 3 outside edges and therefore three orientations for each location.  The number of possible combinations of these corners is (3*8)*(3*7)*...*(3*2)*(3*1) = (3^8)*8!\n\nMultiply the number of possible combinations of edge cubes by the number of possible combinations of corner cubes and you end up with 519,024,039,293,878,300,000 combinations.  \n\nThis is different than the answer I looked up because there are possible combinations when you take the cube apart that cause it to no longer be solvable.  I can't think of an easy way to find the number of combinations that keep solvability.
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Tom baked 4 batches of fairy cakes. Each batch consisted of x cakes. He put 3 cherries on top of each cake in the first batch, but couldn't resist taking off and eating 5 of them. He did the same again with the second batch, then the third, then the fourth! Meanwhile, Sarah baked 5 batches of fairy cakes, again with x cakes in each batch. She put 2 cherries on each cake and an extra two on the final one. She repeated this procedure with each of the batches. She managed to resist eating any at all! Tom and Sarah discovered that Tom's cakes had exactly the same number of cherries in total as Sarah's. Set up an equation for this.\n\nThen: \n\nNumber of cherries on Tom's cakes = 4(3x-5)\nNumber of cherries on Sarah's cakes=5(2x+4)\n\nBut number of cherries is equal so:\n\n4(3x-5)=5(2x+4)\n\nPhew!! I think that works!! Lol!
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5 hour trip times 3 students standing at all time = 900 standing-minutes. 900 min divided by 19 students = 37 min 22 s.\nAt departure student 1 2 and 3 is standing. When a third of student 1's standing-time is over he/she sits down on student 4's seat, who just stood up. When two thirds of student 2's standing-time is over he/she sits down on student 5's seat, who just stood up. Student 3 stands for his fair share of time, and switch place with student 6. All students from #3 to #19 do their standingtime in a single streach. When student 17 is done he/she sits down on student 1's current seat. When student 18 is done he/she sits down on student 2's current seat, and when student 19 is done, they are at their destination!\n\nThe math teacher only has to shout "switch!" every 12min 42sec.\n\nI know you said no trickery, but she could also force the students to all stand for 5 straight hours...
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31 are open.\n\nEach squared number is open at the end: \n1, 4, 9, 16, 25, ... , 961 (31 * 31)\n\nTake any number x.\nList all pairs of factors\neg 6 = 2*3, 1*6.\nIf the factors numbers are different (2<>3, 1<>6) then an even number of students will open and close the locker, so the locker will remain closed. (for six, students 2, 3, 1 and 6 will change the state, 4 students = even = does not change state = closed.)\n\nBut if the factors are the same (it is a square, eg 4 = 2*2) then there is an uneven number of students that will change the state, so the locker will be open.\n4 = 2*2, 1*4, so students 1, 2, and 4 will change the state of the locker so it will stay open.
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Permutation\nSuppose we want to find the number of ways to arrange the three letters in the word CAT in different two-letter groups where CA is different from AC and there are no repeated letters.\n\nBecause order matters, we're finding the number of permutations of size 2 that can be taken from a set of size 3. This is often written 3_P_2. We can list them as:\n\n      CA   CT   AC   AT   TC   TA\n\nCombination\nWhen we want to find the number of combinations of size 2 without repeated letters that can be made from the three letters in the word CAT, order doesn't matter; AT is the same as TA. We can write out the three combinations of size two that can be taken from this set of size three:\n\n      CA   CT   AT
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The answer is 9! / (6!*3!).\n\nYou're dealing with combinations and not permutations, so we don't care which person is which... you really have nine chairs, six of which are occupied and three of which are empty. Forget about ABCDEF being the six people: Let X represent an occupied chair, and O an empty chair.\n\nIf you need to create a list of all possible combinations, just start with the X'es all to the left and the O's to the right, then do a bubble sort thingy.\n\nXXXXXXOOO\nXXXXXOXOO\nXXXXXOOXO\nXXXXXOOOX\nXXXXOXXOO\nXXXXOXOXO\nXXXXOXOOX\nXXXXOOXXO\nXXXXOOXOX\nXXXXOOOXX\netc.\n\nIf you grind your way through the whole list, you'll find there are 84 combinations. There's some neat stuff going on here with triangular numbers: you can see that there's one arrangement where the first O is in the 7th position, three where the first O is in the 6th position, six where the first O is in the 5th position... follow the pattern and you'll see that there are 1+3+6+10+15+21+28 combinations in the list. That adds up to 84.\n\nHopefully that helps!
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I was able to draw 16 different forms (8 with their symmetrical reversed forms).  There may be more but I don't have the time to work on it.\n\nI took another look at the problem and realized that in addition to the left-right symmetry there is also a top-bottom symmetry that adds at least another 14 answers to the 16 that I came up with earlier.  So there are at least 30.\n\nCould be more, but it is my bed time.
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1. This will waste a lot of space, but i will go step by step and try to explain. First, write down all the adults and children (10 adults, 15 children) and the number of kids in each family\n\nAdults\nBrad\nRobert (2 daughters, 3 sons)\nLenny\nBen (son=Sam)\nRick/Leslie\nGayle\nJenny\nHolly\nVickie (1 son, 1 daughter)\n\nChildren\n\nGIRLS\nBarb (laura's sis)\nGina\nJill\nLisa Scorps (3 brothers)\nLaura (barb's sis)\n\nBOYS\nBilly (Scott's bro)\nBob (no brothers)\nJack\nJohn\nLuke\nRon Vernon (no brothers)\nSam\nScott (Billy's bro)\nTim Baldwin\nWayne\n\n2. Since Rick and Leslie have children with identical initials, if you go through the alphabet, the only letter is J. (cant be S or B because Scott and Billy are brothers, cant be L because Laura is Barb's sister and cant be anything else because there arent 2 or more with same starting letter) so that means that Rick and Leslie's children are some kinda combination of Jack Jill and John. But, they can't have 2 kids since Vickie has 2 kids, so that means that they have 3.\nAlso, Gayle has more kids than Jenny, who has more than Holly. This has to mean that Gayle has 5 kids and is married to Robert (since Robert also has 5 kids), and Jenny has 4, and Holly has 1. (since 2 and 3 are gone by Vickie and Leslie)\n\nAdults\nBrad\nLenny\nBen (son=Sam)\nRobert/Gayle (2 daughters, 3 sons)\nRick/Leslie (3 kids = 1 daughter, 2 sons = Jack, Jill, John)\nJenny (4 kids)\nHolly (1 kid)\nVickie (1 son, 1 daughter)\n\nChildren\n\nGIRLS\nBarb (laura's sis)\nGina\nJill\nLisa Scorps (3 brothers)\nLaura (barb's sis)\n\nBOYS\nBilly (Scott's bro)\nBob (no brothers)\nJack\nJohn\nLuke\nRon Vernon (no brothers)\nSam\nScott (Billy's bro)\nTim Baldwin\nWayne\n\n3. Since there are only 5 girls, and from the list, Robert/Gayle have 2, Rick/Leslie have 1 and Vickie has 1. this means that out of Holly and Jenny, they have 1 girl kid. and since Barb and Laura are sisters, they have to go to the family with 2 daughters, which is Robert/Gayle. \nLisa Scorps has 3 brothers and she isnt in the family with 5 kids, so she has to be in the family with 4 kids. This means Holly's kid is a son. \nBob and Ron Vernon have no brothers, which means they have to be sons of Holly and Vickie. Ben's son is Sam, and since Sam can not be in either Holly or Vickie's family, he has to be Jenny Scorps' son and that means Ben is married to Jenny Scorps.\nSince Vickie isn't married to Lenny, she has to be married to Brad and Lenny has to be married to Holly.\n\nAdults\nRobert/Gayle (2 daughters, 3 sons = Barb, Laura)\nRick/Leslie (1 daughter, 2 sons = Jack, Jill, John)\nBen/Jenny Scorps (1 daughter, 3 sons = Lisa, Sam)\nHolly/Lenny (1 son = Ron or Bob)\nVickie/Brad (1 daughter, 1 son = Ron or Bob)\n\nChildren\n\nGIRLS\nBarb (laura's sis)\nGina\nJill\nLisa Scorps (3 brothers)\nLaura (barb's sis)\n\nBOYS\nBilly (Scott's bro)\nBob (no brothers)\nJack\nJohn\nLuke\nRon Vernon (no brothers)\nSam\nScott (Billy's bro)\nTim Baldwin\nWayne\n\n4. Now that you have all the couples, you have to start finding their kids and last names. Since Wayne is not related to Sam, he has to be in Robert/Gayle's family (since the other families dont have more sons).\nHolly isn't Mrs. Vernon, which means she cant be Ron's mom.\nNow, it says Tim Baldwin is one child. He has to be in either Ben/Jenny's family or Robert/Gayle's because they are the only ones who dont have enough sons. But since his last name is Baldwin, he cant be in Ben/Jenny's because their last name is Scorps. So he is in Robert/Gayle's family.\nScott and Billy are brothers, and the only family that has 2 unknown sons is Ben/Jenny.\nRick and Leslie are not the Webers, so they have to be the Cannons and Holly/Lenny have to be the Webers.\nNow you look at the only girl kid left and that has to be the daughter of the Vernons. The only guy kid left has to be the son of the Baldwins. They are