Four couples have reserved seats n a given row for a concert. In how many different ways can they be seated if the two members of each couple wish to sit together?\n(The actual answer is 384 but how!!!!!!!)The answer came from the back of the book!
134384
First think of a situation involving 4 seats for 4 people:\n\nIn the first chair any of the 4 can sit, in the second, any of the remaining 3, in the third either of the remaining 2 and the last person gets the 4th seat\n\nthe number of combinations is 4*3*2*1= 24\n\nThis is the same for 4 couples since each couple sits together.\nNext, each couple can be in one of two positions because the members of a couple may switch seats and still be sitting together.\n\nthis multiplies the original number of combinations by 2^4 or 16\n\n24*16=384
77842
In the experiment of choosing a committee of 6 from a group of 8 women and 7 men,\n\nTotal no. of possible choices \n= no. of ways in which the 6 members of the committee can be chosen from the total 15\n==> n = C(15,6)\n==> n = 15*14*13*12*11*10/6*5*4*3*2*1\n==> n = 7*13*11*5\n==> n = 5005\n\nLet "A" be the event of choosing the committee with 3 women and 3 men\n\nFor Event "A"\n\n---------- Men----Women---Total\nAvailable---8-------7------15\nTo Choose---3-------3-------6\nChoices----C(8,3)-C(7,3)--C(15,6)\n\nNo. of favorable choices\n= (no. of ways in which the 3 men can be chosen from the total 8) * (no. of ways in which the 3 wommen can be chosen from the total 7) \n==> m(a) = C(8,3) * C(7,3)\n==> m(a) = 8*7*6/3*2*1 * 7*6*5/3*2*1\n==> m(a) = 8*7 * 7*5\n==> m(a) = 1960\n\nProbabilty of choosing the committe with 3 women and 3 men \n==> Probability of occurance of event "A" = m(a)/n\n==> P(A) = 1960/5005\n==> P(A) = 392/1001\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/
415320
MSU's seating info-\n\nSection F has 59 rows\nThere are 44 seats in the row and they are numbered from low to high, working towards the endzone.\n\nHere is a view from that section- http://www.msufieldview.com/section_f.html\n\nI hope this answers your question.
378800
i.  Since order matters, this is 250P14, the permutation of 250 days taken 14 at a time.  (This is a huge number.)\n\nii.  No; they could all have 3 Mondays, 3 Tuesdays, 3 Wednesdays, 3 Thursdays, and 2 Fridays.\n\niii. 20x14 = 280 total days which means on at least 30 days, two people are off and at most 220 days, only one is off.  It seems like on those 30 days, since there are 20 workers, there are plenty of ways to make 30 pairs without repeating the same pair.  I hope this is right!
361031
Oh Dear! lots of dodgy maths out there!\nWelcome to Mr Cantor's infinity paradise hotel!\nWe got infinite rooms, but they're all full just now.\nDon't worry!.. we'll squeeze you in!..I'll just get every guest to move one room along\nNow you can get in room1\nHey there's an infinite stretch limo arrived with an infinite number of guests..no probs just get all our guests to move from room n to 2n!\nYes infinity is an uncountable number, but you can still do mathematical ops with it\nHope that helps, you can work out the rest foor yourselves. PS always agree with your wife- she's right even when she's wrong-now that's a true paradox
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.
101274
I can't count, so don't blame me if I get this wrong.\nLet n be the number of participants.\nSince it's a close circuit, the people in front of him are in a sense also behind him so there ara (n-1) people both ahead and behind him.\nso (1/5+5/6)(n-1)=n\n31/30*n -31/30=n\n1/30*n=31/30\nthus n=31\n31 total people
546155
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.
518118
Try to factorise number 319.\nIt is 11 * 29\nSo you have 11 rows 29 people per row \nor 29 rows with 11 people per row.\n\nIf you wonder how to find it it is simple\nTry to divide 319 with prime numbers\n\n2,3,5,7,11,13,17\n\nOnly 11 is ok!
666229
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.
125073
Querier's answer to the first part was excellent.  Here's the second part, which is even simpler.  Suppose you've chosen the married couple.  That gets you 1 man and 1 women.  Then there are 7 ways to choose the 3rd person.  So 7 of the 70 committees from part one include the married couple, leaving 63 that do not.
67839
"An orchestra consists of string instruments and wind instruments in a ration of 9 to 4. There are 91 string and wind instruments in the orchestra. How many string instruments are there? How many wind instruments?"\n\nString Number = X\nWind Number = Y\n\nx + y = 91\n9x = 4y  --> y = 9/4x\n\nx + 9/4x = 91\n13/4x = 91\n\nx= (91*4)/13  --> x= 28\nthere you have the number of strings.\nnow you refer to the second equation (9x = 4y)\nand substitute X\n\n9(28) = 4y\n252 = 4y --> 252/4 = y = 63\n\nyou have 28 strings and 63 winds that added together are 91\nand if you calculate, every nine strings you have four winds.\n\nTen points here?
161474
Arrangements: 9!/5!2!=1512. See Ron's answer for explanation.\n\nSelections: 12\nS and E are special, since there are more than one of each.\nSo we count each case seperately:\n4S        1\n3S0E      2\n3S1E      1\n2S2E      1\n2S1E      2\n2S0E      1\n1S2E      2\n1S1E      1\n1S0E      0\n0S2E      1\n0S1E      0\n0S0E      0\n\nAdd the numbers up, and we have 12 selections.
360189
1024 matches. This should be on the mathematic section. Everyone know that an elimination match would always require <number of player> - 1 to finish. \n\nThe simple logic is actually: We're gonna create 1024 losers, how could we create 1024 losers with less than 1024 match? Can't have 1024 losers with only several matches.\n\nEach player would play the match 10 times (2^10 = 1024), and exactly one player would play 11 matches. (to complete up the 1025 players)
440430
Let x = number of ride tickets.\n60 = 3x + 12\nSubtract 12 from each side\n48 = 3x\nDivide both sides by 3\n16 = x\nAnswer: 16 ride tickets.\n\nCheck.\n16 tickets at $3 each would cost\n16*3 = $48\nplus the $12 in refreshments\n48+12 = $60
625585
Both John ad his wife shake hands with 4 people.\n\nI'll try the briefest explaination I can.\nThere are nine answers 0-8 plus Johns which is unknown for a total of 10.\nTake any couple for an example.  Which couple won't matter because we will figure the total for one partner which will give us the answer for the other partner.  Once we have accounted for each number of handshakes, the missing number of handshakes will be the unaccounted for partner - John, and of course the one remaining known value will be his wife.\n\nRemember every person will surely not shake their own hand or their spouses.\nOne person shakes no hands. He/She knows everyone.  That means everyone knows him/her and will also not shake his/her hand.  You now have 4 couples who will not shake their own, their partners, or his/her hand.  That is 8 people who will shake at most, 7 hands.  The only person who could possibly shake 8 hands will be that persons spouse.  So couple one shakes 0 and 8 hands.\npick any of the remaining 4 couples.  Any single person of the remaining 8 must shake the hands of at least one person. (because we have 1-7 remaining) If he/she is unknown to one person, then that means he/she is known to 6 of the seven remaining persons. One of which is his/her spouse.  So there are 6 people (the 3 other couples who know and won't shake hands with themselves, their spouse, this individual or the first person who shook hands with no-one) who will not shake hands with more than 6 people.  That makes this persons spouse the one that shakes hands with 7 people. The spouse knows the first person, themselves and the spouse.\nAnyways the other two couples work out to shaking 2 and 6 hands, and shaking 3 and 5 hands.  The remaining person shook 4 hands (the last number) and is coupleless/without a spouse so that person must be Johns wife and she and John both shook 4 hands.  \n \nBy couple: 0/8, 1/7, 2/6, 3/5, 4/4.   We know 0,1,2,3,4,5,6,7,8 were given as answers at the party so the only single (non couple) is 4.
439200
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!