3 men and 3 women are sitting at a round table , each woman being flanked by two men and vice versa. How  many different seating arrangements are possible such that in no arrangement,every man is flanked by the same women ?\n\na)12\nb)720\nc)6\nd)120\n\n\nI know that if there is n person sits on a round table....then sitting could be arranged in (n-1)! ways.\n\nbut in this problem,they have restrcted the sitting arrangements by saying "... no arrangement,every man is flanked by the same women...".\n\nhow do i solve then ?\n\n\nRegards
125798
well if the men are abc and women are def, then  the arrangement is like adbecf  right?  so   man a is "flanked" by d and f, and man a could also be "flanked" by de or ef so theres 3 different ways man a could sit. for each of these, there are 2 positions the other 2 men could sit in. \n\nso basically its 1 of these:\nadbecf\nadcebf\naebdcf\naecdbf\nafbdce\nafcdbe    \nit doesnt matter which seats they sit on, cuz that will make each man be "flanked" with the same women so u cant have that. all that matters is their order.\nthe answer is c. i hope this makes sense.
96829
For the first question, that means you want to know about committees with exactly 3 women and committees with exactly 4 women. Since these are independent cases, we consider each separately and add the result. \nFor 3 women, we need to know how many unordered groups of 3 women we can choose from 4 women, so we have 4[Choose]3. The men have 2 people on this committee, so they contribute 4[Choose]2 to the product for a total of (4[Choose]3)*(4[Choose]2) committees with exactly 3 women on board. \nThere is only one way to grab an unordered set of 4 objects from a set of 4 objects, so we go right to the men. There are 4 ways the men can fill the last spot, so this is 1*4=4 different committees with 4 women on board.\nThus the total number of committees with a majority of women on board is (4[Choose]3)*(4[Choose]2) + 4.\nThe second part is very similar to the first.
268497
hmmm... looks like u guys need one more guy. count me in. that way 3 guy and three girl, so one will be left out.
324033
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
361219
This is a very difficult counting problem.  I will show one possible grouping and count the number of ways it can be done.\n\nTo put just 1 boy in group 1, there are 10 ways to pick the boy and "combination of 15 things taken 4 at a time" ways to pick the girls.  This is 10 * (15*14*13*12 / 1*2*3*4) = 10*1365 = 13650 ways to pick 4 girls and 1 boy.\n-----------------\nThere are many grouping scenarios; let us start with 1 boy, 1 boy, 1 boy, 2 boys, 5 boys.\n\nWe know there are 13650 ways to pick a single boy for group 1.\n\nTo then pick a single boy for group 2 from the remaining children, there are 9*(11*10*9*8 / 1*2*3*4) = 9 * 330 = 2970 ways.\n\nIf a single boy is then selected for group 3, there are 8 * (7*6*5*4 / 1*2*3*4) = 8*35 = 280 ways.\n\nGroup 4 must have at least 2 boys.  There are 7 boys left, and we must choose 2 of them.  So there are 7*6/1*2 = 21 ways to choose those boys with the remaining girls.\n\nThe last group would have 5 boys.\n\nSo just for this one possible way of grouping - 1 boy, 1 boy, 1 boy, 2 boys, 5 boys, there are \n13650 * 2970 * 280 * 21 * 1 = 238,378,140,000 ways to arrange them.\n\nTHIS MAY NOT BE THE CORRECT FIGURE FOR THIS GROUPING SCENARIO - THERE MAY BE DUPLICATES, IN WHICH CASE THIS FIGURE IS TOO HIGH.  I did not know how to remove them from the figure.  The 13650 is correct for the first, but then we need to look at how many ways are there to grab 4&1, 4&1, 3&2, 0&5 from the remaining children.\n\n\nWe must also account for other grouping scenarios, such as 2, 2, 2, 2, 2 boys; 1,1,2,2,4 boys; 1,1,1,3,4 boys; 1,2,2,2,3 boys, and some others.  For each of these grouping possibilities, the above such calculations would have to be done.  The sum of all the grouping possibilities is the total.\n\nSo this is a very difficult counting problem.  I would think you should search for more tools from your professor/teacher, or book.  There may be some more advanced way of saving time rather than go through each grouping scenario one at a time; though that will give the correct answer.
661365
In all these problems you are going to pick a set of objects or perform a set of actions.  \n\nIf the ordering of the outcomes makes a difference, then it's a permutation. If the ordering doesn't make a difference, then it's a combination.\n\nExample 1: The 10 runners run a race. They give medals for the first 3 places. How many different outcomes are there?\n\nThe order of finish makes a difference, since it determines who gets which medal, so it's a permutation problem.\n\nAnswer: 10x9x8 = 720\n\nExample 2: Three students are chosen from a class of 10 to clean after a food fight. How many ways can the group be chosen?\n\nIt doesn't matter who gets picked first, they all have to clean. Ordering doesn't make a difference, so it's a combination problem.\n\nAnswer: 10C3 = 10!/3!7! = 10x9x8/3x2x1 = 120.
647018
1) Let's first consider not the combination but the permutations without repetition (considering the order).\n\nSuppose to have 39 numbered cards in your hands and shuffle them well. \nIf you get one card and you drop it on the table you have 39 different possibilities, obviously.\n\nNow, from the remaining 38 cards in your hands, If you drop another card on the right side of the previous card, you'll have 38 different possibilities for just that second choose, but If you consider the order of both cards you'll have exactly 39*38 = 1482 different possibilities that the 2 card will be ordered and choosen in that way.\n\nIf you get a 3rd card from the remaining 37 in your hand you'll have on the table in the right side of previous exraction, a group of 3 ordered cards choosen by 39*38*37 = 54834 different ways will be on the table.\n\nRecursively go to choose until 6 ordered cards will be the table and you'll have 39*38*37*36*35*36 = 2349088560 possibile ways if you consider the order. \n\n2) Now let's see what happend when the order of the selection matters as you asked in your question. \nSo let's make this consideration:\n\na) If you have one card (A) you don't mind about ordering problems:  {A} = (A)\n\nb) if you have two cards (A, B) you have to consider the couple (A, B) = (B, A) so we write {A, B} are commutative;\n\nc) if you have three cards "A", "B" and "C" you can group (A, B) as one element {A, B} because order doesn't matter (point b).\n\nSo we can consider the 3 cards (A, B, C) as:\n ( {A, B}, C ) = ( C, {A, B} ) as we made in the point b.\nYou could also group (A, {B, C}) = ( {B, C}, A)  or {A, C} and then swapping with B.\n\nSo with 3 elements we have 3 different ways to create couple to swap so we can create 3*2 = 6 different combination.\n\nIf we have 4 cards we know it's possible to choose 4 different ways to create list of 3 cards. Let's stop one each time with your finger and you can see that you can group easily the remaining other 3's.\nSo the possible combination with 4 cards is 4 times the possible combination I have with 3 cards. So it's  4 * (3 * 2).\n\nWith 5 cards you'll have 5 times the number of combinations we found with 4 cards, so it's  5 * (4 * 3 * 2).\n\nFinally, with 6 cards in a table arrange in 6 * 5 * 4 * 3 * 2 = 720 different ordered ways.\n\nNow, your question was how to do if the order of selection matters and in the previous example we find that the number of possibilities we have when we select 6 cards ordered is this number 39*38*37*36*35*34 = 2349088560. And we know now that this number is 720 times greater than the possibilities wether the order of selection doesn't matter. (720 = 6 * 5 * 4 * 3 * 2).\n\nIn other words, 39 * 38 * 37 * 36 * 35 * 34 is the number of permutations without repetition of 39 elements grouped by a list of 6 elements considering the order.\n\nAnd now we know with 6 elements you can have 6*5*4*3*2 possible different positions of those elements. \n\nSo Just divide 39*38*37*36*35*34  by 6*5*4*3*2 and you'll have the result of the combinations.\n\nSo to see how many combination of 6 cards bye a set of 39 just divide 39*38*37*36*35*34 / 6*5*4*3*2 = 3262623.\n\nthe number 6*5*4*3*2 could be wrote "6!" using the the factorial function.\n\nThe factorial function of positive integer n, is defined as the product of first n numbers:\n\nn! = 1 * 2 * 3 * ... * n\n\nand is a convention that the product of no numbers at all is 1.\n\n0! = 1\n\n\nThe factorial is also defined recursively by the follow definition:\n\n(n+1)! = (n+1) * n!\n\n\nSo if you have n object and you want to know how many permutation (the order matters) without repetition (each object can be chosen only once) the objects makes with a number k of selection the formula is:\n \n  n! / (n-k)!\n\nso it's the product of k terms starting from n and descending until reached (n-k+1).\n\nn*(n-1)*(n-2)*.
72805
there are 6 place for 15 student so:-\n\n15*14*13*12*11*10=answer\n\nor simply use combination since the arrangement is not important\n\n15C6
359881
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...
164947
if you think about it there are 8/20 woman if you put 8/20=X/30 and cross multiply then you will get  20X=240 divid both sides by 20 and you will get X=12 that means theere are 12 woman in the class and then if you 30-12=18\nso there are \n\n        12 woman and 18 men in the otherclass
559964
You place all 9 squares. There are 9! ways to place the pieces(without being concerned about each pieces orientation)\nFor each of the 9!, each piece can be rotated into 9 positions. So there are going to be 4^9 possiblities.\n\nOnce you find the correct solution, you can rotate the entire puzzle 90 degrees 4 times.\n\nso the correct number is 9!* 4^9 /4\n9!*4^8\n= 23,781,703,680\n\nNOTE--To the people who are answering 9!*4^9\nWhat would the possible solutions be if there was only one piece?\nWould it be 1!*4^1 =4\nNo, it would be 1/4 of that.\n\nIf there is a similar puzzle of NxN size\nthe solution would be (N^2)! *4^(N^2 -1)
505114
If 4/7 are male, then 3/7 are female.\n\nLet x be the total number of students:\n(3/7)x = 3,600\n\nMultiply both sides by 7/3:\nx = 3,600 * 7/3\n\nSimplify:\nx = 1,200 * 7\nx = 8,400 students total
54139
That's an extremely good question. \n\nThe answer is simple. Before that, i will explain what your question means to those who don't get. Well, it means that we must get 24 using 1,3,4 and 6 any number of times however involving operations +,-,* and / only once. \n\n  Then, the answer is very simple :\n\n  {[(6*6)/(3-1)]+[6]}  =  24\n\n---------------------------------------------------------\nWOKRING :\n\n {[(6*6)/(3-1)]+[6]}   =  {[(36)/(2)]+[6]}\n                       =  18 + 6\n                       =  24\n\n----------------------------------------------------------\n\nI hope you post more questions like these. They evoke a lot of thought. I hope the solution serves th purpose.
269571
Let's simplify the problem.  What if the question was how many people must attend a party to guarantee that a group of 2 people that all know or don't know each other attends.\n\nThis answer is obviously 2.  They either know or don't know each other so you are done.\n\nNow let's try the case with figuring out how many people must attend such that you have 3 people that either know or don't know each other.  You can start with your two people.  Let's say they know each other.  Then...\n\nYou know, I'm going to have to think on this a little longer.  It's not as straightforward as saying it is 5 or 17, but requires some recursive logic...  I'm sure there is a series that explains the relationship.  Let me go look and get back to you.\n\nOkay, I found an article on Wikipedia covering this graph problem.  Let's go back to the case of 3 relationships.  In this case, you can have 5 people you can color the relationship graph with two colors (red = know, blue = don't know).  You can color this without creating a triangle of one color.  However if you try to add one more node to the graph (guest), there is no way to prevent a triangle.  So for 3 relationships (all known or unknown) the answer is 6 guests.\n\nFor 4 people, the answer is 18 guests are required to guarantee this.\n\nHowever for 5 people, the answer is between 43 and 49, but no one has proven the lower bound.  If you had 49, you would definitely guarantee this condition.  But it might be that you could also do it with 48, 47, 46, etc.\n\nSo the answer is somewhere between 43 and 49 people.
383443
26. The pattern has to start 1, 2, 4, 8, 15. I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3. The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.\n\nAs for why this makes sense: Consider the question in two dimensions, i.e. into how many parts will many random lines diivide a plane?  That pattern starts 1, 2, 4, 7, 11 - the same numbers on the differences line above.  If you want to get really abstruse, I would submit that the number parts random spaces divide the fourth dimension is 1, 2, 4, 8, 16, 31, 57.  :)\n\n* * * * * *\n\nHey, you really do ask an interesting question, but it appears I'm the only one willing to answer it.  (Yes, I see you asked the same question a few days ago.)  OK, then, let's look at it more systematically:\n\n\nOne-dimensional question: Into how many parts will random points divide a line?\n\nEasy question, obviously:  Zero points leave one part (i.e. the whole line); one point breaks it into two parts, two points into three parts, etc.  The sequence is 1, 2, 3, 4, 5, etc.  Generally, n random points divide a line into n + 1 parts. \n\n\nTwo-dimensional question: Into how many parts will random lines divide a plane?\n\nHarder question, but you can work this one out on paper.  Zero lines leave one part (the whole plane); one line breaks it in two; two lines breaks it in four.  Then it gets tricky - the third line can't go thru all four parts, but it can go through three of them, so you end up with 7 parts.  The fourth line can go through 4 of the 7 parts.  Now, how you'd prove this rigorously I don't know, but the sequence is 1, 2, 4, 7, 11, 16, etc. and is clearly derived from the one-dimensional sequence: start with 1, then add 1 to get 2, then add 2 to get 4, then add 3 to get 7, then add 4 to get 11, etc.  To put this into a formula, n random lines divide a plane into (n^2 + n + 2)/2 parts.        \n\n\nNow for the three-dimensional question: Into how many parts will random planes divide a three-dimensional space?\n\nTo visualize this one, look at a corner of your room.  Obviously, zero planes leave one part (the whole space), one plane  breaks it in two (visualize the floor of your room as the plane, for example), two planes breaks it in four (visualize the floor plus one wall), and three planes breaks it in eight (visualize the floor plus two adjoining walls).  Now, try to imagine the fourth plane - it can't go through all eight parts, but it can go through seven of them, making a total of 15 parts.  The three-dimensional  sequence starts 1, 2, 4, 8, 15, and again we see it is derived from the sequence for the previous dimension: start with 1, then add 1 to get 2, then add 2 to get 4, then add 4 to get 8, then add 7 to get 15, etc.  To get the number for 5 planes, add 11 to get 26.  To put this into a formula, n random planes divide space into (n^3 + 5n + 6)/6 parts.   \n\nBased on what we are seeing, what the heck - why not carry it forward to the fourth dimension.  Using the numbers from the three-dimensional sequence, the sequence becomes 1, 2, 4, 8, 16, 31, 57, etc.  We can conclude, for example, that 6 random three-dimensional spaces divide the fourth dimension in 57 parts.  Now that should really impress your instructor!
412851
My opinion is that no one will be able to answer your 'question' because it doesn't make sense in any known language.\n\nHowever, I will try.  Imagine a universe that is divided into two large groups.  \n\nThe members in group A include all 14 of your conditions and some others as well (e.g., 15 blue, 16 tall, etc).\n\nThe members of group B includes everything not included in group A.  \n\nItems 1 to 14 'may' be represented in some subsets of group A.  (Paragraph 1)\n\nNo members of group B will ever be represented in any subset of group A.  (Paragraph 2).
571832
superobotz is right, and i dont mean to patronanize or act smart or anything, but i just want to explain something about why certain steps have to round like they do, just in case it wasnt making enough sense  :)\n\nthere can only be 10 in a stack because . . .\n   you divide the max height (10 m) by the height of each desk (.95 m) and get that each stack can have about 10.53 desks.  now obviously, since you cant put part of a desk in a stack you cant have any more than 10\n\nyou have to round up at the end because . . .\n   208 desks divided by 10 desks per stack gives you 20.8 stacks, so you will have 20 full stacks and 1 shorter stack, but the short stack still has to count as a stack so its 21.