A tourist wants to visit 3 or more of the 5 major cities in india : Delhi,Mumbai,Chennai,Kolkata, and Bangalore.\nIn how many ways  can he plan his tour such that Delhi is always included ?(No city is repeated,the sequence of cities is relevant for a tour.)\n\na)5(4!+3!+2!)  \nb)5!+4*4!+6*3!\nc)5!+4!+3!\nd)5C5+4C3+4C2\n\nwhich is the answer ?
124549
First.\n"the sequence of cities is relevant for a tour"\nmeans order is important. Thus, a permutation reflects each different tour.\n\nThere are 3 possibilities for the number of cities included for the tour, namely, 3,4, or 5 cities. We consider each case separately.\n\nCase 1. ( 3 cities )\n\nDelhi is included. The remaining 2 cities can be chosen from the remaining 4 cities in 4C2 ways = 6 ways.\n\nAfter the 3 cities are selected, they can be permutated in \n3P3 ways = 3! ways. Each representing a tour.\n\nHence, Total number of tours in case 1 = 6 x 3! ways.\n\n\nCase 2. ( 4 cities )\n\nDelhi is included. The remaining 3 cities can be chosen from the remaining 4 cities in 4C3 ways = 4 ways.\n\nAfter the 4 cities are selected, they can be permutated in \n4P4 ways = 4! ways. Each representing a tour.\n\nHence, Total number of tours in case 2 = 4 x 4! ways.\n\nCase 3. ( 5 cities )\n\nDelhi is included. The remaining 4 cities can be chosen from the remaining 4 cities in 4C4 ways = 1 way.\n\nAfter the 5 cities are selected, they can be permutated in \n5P5 ways = 5! ways. Each representing a tour.\n\nHence, Total number of tours in case 3 = 1 x 5! ways.\n\nThus,\n\nTotal number of \n"major city tour is" \n\n= "no.tour in case 1,2 and 3."\n\n= 6 x 3! + 4 x 4! + 5!\n\n= 5!+4*4!+6*3! ways.\n\nThe answer is (b)\n\nRegards
95348
Changed my mind folks. I thought about the actual question and realised it should be a combination question (geez we can tell who didn't do maths at school - me!)\n\nThe answer is\n\n10 x 9 x 8 = 720 divided by 3 x 2 x 1 = 6\n\nMakes the answer 120.\n\nLet me explainacate - (new word for you there)\n\nThere are ten cd's to choose from when you pull the first, 9 to choose from when you pull the second and 8 to choose from when you select the third.\n\nThen you divide it by 6, because the 3 CD's you did pull out could be listed in 6 different permutations, but they are still the same CD's\neg:\nabc, acb,bac, bca, cab, cba - listed different, but same content.\n\nTherefore, I agree with Tanny - It is a combination question and the answer is 120 for the reason outlined above.\n\n(excuse my previous idiot-ness) (another new word there folks)
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The first letter can be any one of the 5 letters\nThe second one can be any of the 4 remaining letters\nand so on\n\nSo the answer is 5 x 4 x 3 x 2 x 1 = 120 different ways\n\nIf two of the letters are the same (eg green) then divide by (2 x 1) to give 60\n\nIf three of the letters are the same (eg eerie) then divide by (3 x 2 x 1) to give 20\n\nIf there are two pairs of letters (eg yuppy) then divide by (2 x 1) x (2 x 1) to give 30
384951
1) \n\n1+2+3+4+5+....n +(n-1)+(n-2)+...+3+2+1...regrouping\n\n=2(1+2+3+4+5+....(n-1)) +n...regrouping the term in parenthesis\n\n=2( (1+n-1)+(2+n-2)+(3+n-3)...)+n\n=2(n+n+n+...(n-1) terms)+n\n=2(n)(n-1)/2 +n\n=n^2-n+n\n=n^2\n\n\n2)\n 1=10^0 and 11=(10^1+10^0) and 111=(10^2+10^1+10^0)...\n\n(10^n+10^(n-1)+....10^2+10^1+10^0) times itself\n=(10^0+10^2+10^3+....10^n)^2\n=10^2n+2*10^(2n-1)+.....\n\n\n3)\n\nn^2-n=n(n-1)...one of them must be even and is therefore divisible by two.\n\nn^3-n=n(n^2-1)=n(n+1)(n-1)=(n-1)(n)(n+1) or three consecutive integers.\n\nIf the product is three consecutive integers, then one of them must be divisible by three and therefore the product is divisible by three.
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You are placing the letters MMMIINU into seven boxes, one per box.\n\nThe Ms can go in C(7,3) = 7!/(3!4!) = 35 ways.\n\nThere are four boxes left, and there are C(4,2)=6 ways to put choose two locations for the Is.\n\nThen there are two boxes left, and exactly 2 ways to choose where the N and U go.\n\nSo the total is 35*6*2=420.
603235
To get your answers, imagine what has to happen.  \n\na) First, you have to pick 1 of the 4 white balls out of the 12 balls there.\n\n2nd, you have to pick 1 of the 3 white balls left out of the 11 balls there.\n\nSo, in order to get a red and then a red, you need to get 4 out of 12 followed by 3 out of 11 or 4/12 times 3/11 or 12/132 = 1/22.\n\nb) First, how many ways can 5 people line up?  5!.  How many ways are correct (assuming they all have different names, only 1).  So, the prob is 1/5! or 1/120.
460869
1) 4-3x<_8+x\n4+8<_x+3x\n12<_4x\n3<_x or x in the interval [3,plus infinity)\n\n2) 3x-3>-x+1\n3x+x>1+3\n4x>4\nx>1 or x in the interval (1, plus infinity)\n\n3) 3(x-1)_<4x\n3x-3<_4x\n-3<_x or x in the interval [-3, plus infinity)\n\n4)1/2x<x+1\n-1<x/2\n-2<x or x in the interval (-2, plus infinity)\n\n5)-1/2x>1-x\nx/2>1\nx>2 or x in the interval (2, plus infinity)
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1.) if I am understanding this correctly that is an easy one: just start with 2 and go up by 10\n\n2, 12, 22, 32, 42...\n\n2.) the pattern for this is adding 8 more to the previous add \n\n15-3 = 12\n35-15=20\n63-35=28\n\nsee how its adding 8\n\nso the 17th number would be: 1155 and the 37th number would be: 5607
234093
I think it should have been C for Charlie, but if A is Charlie then here are the events.\n\nCharlie chosen ist:\nABC ABD ABE ACD ACE ADE ACB ADB AEB ADC AEC AED\nCharlie chosen 2nd\nBAC BAD BAE CAD CAE DAE CAB DAB EAB DAC EAC EAD\nCharlie chosen 3rd\nBCA BDA BEA CDA CEA EDA CBA DBA EBA DCA ECA DEA\n\nTotal events 36. This is correct since the the number of distinct possibilities is 5x4x3 = 60\nAnd the no of ways he won't be chosen are 4x3x2=24\n60-24=36
339724
Answers choices are:\n     1, 4, 3, 2, 2,\n      3, 1, 4, 2, 1,\n      1, 1, 4, 2, 1\n      3, 2, 3, 4, 1\n      3, 4, 1, 2, 4
62398
Problem 1:\n\nSITA = 12 WINS\n\nGITA = 20 WINS (SINCE SITA HAS TO GIVE HER 8 CHOCO AT THE  \n       END OF ALL GAMES)\n\nTOTAL = 32 GAMES = 32 DAYS\n\nTHEREFORE, SITA STAYED 32 DAYS AT GITA'S HOUSE.\n\nProblem 2:  \n\nMMMMM...@#$%&*...ZZZZZZZ....SNORE....????\n\nProblem 3\n\nC - is the reference point. He is in third place.\n\nA - he can either be in 1st or 4th place only since he is \n  to be followed by E. If he is in 1st place that would \n  make him a liar which means that he should not be in 1st \n  place. Therefore, he must be in 4th place.\n\nE - follows A, therefore, he is in 5th place.\n\nB -is in 1st place (gold). Liar. So he must be in 2nd place.\n\nD - is not in 1st place (silver). Another liar. So he must be in 1st place. Also, that is the only remaining place.\n\nSo the correct order (From 1st place to 5th place):\n\nD - B - C - A - E
261466
I saw this before..a long time ago. The answer is 2, 2 and 9.\n\n1) Break down 36 into prime numbers - 36 = 2*2*3*3*1\n\n2) Now, we must combine these into ages. Three of them. Any of the ages can be 1, but the others must still come to 36 when multiplied. We also know that when added up, they must be equal to the bus number.\n\nPOSSIBLE COMBINATIONS (Product of ages, Sum of ages)\n\n1, 1, 36  = 38\n1, 2, 18 = 21\n1, 3, 12 = 16\n1, 4, 9 = 14\n1, 6, 6 = 13\n2, 2, 9 = 13\n2, 3, 6 = 11\n3, 3, 4 = 10\n\n3) The number that they add up to is a number the friend knows - the bus number.YET he still says asks for more data! That means that there must be more than one occurrence of the sum in the addition answers. Otherwise, he would already know their ages. Therefore, the only sum that appears more than once is 13, with the sets {1, 6, 6} and {2, 2, 9}. The answer must be one of these.\n\n4) Your clue said "The older one has blue eyes". This means that must be ONLY ONE older child. Therefore, leading us to the answer of the kids' ages to be 2, 2 and 9.
128393
well we know the 3 space must be even, in order to avoid the last space resulting in another 5\n\nhmmm... i'l work on it, provided you're working as well!\n\n..and #3 mustn't be 0, to avoid another 0\n..lol i'd have had it if i had been home...i made a program to do it, and then left before i finished/ran it lol\n\nbut that, is of course, the correct asnwer, 396*45=17820\n\nhaha thanx for the challenge!
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Tim is right.  i just want to explain it further so you will understand.\n\nso the problem is we are looking for the time spent in both trips. lets call it "X". one is 1 hour longer than the other so we can call it "X + 1". \n\nif trip1 with a distance of 180 miles is taken in X with an additional 1 hour we can call this equation as 180 per X+1 or in mathematical form its 180/(X+1)\n\nif trip 2 with a distance of 135 miles is taken in X we can call it 135 per X or 135/X\n\nwe have now 2 equations 180/X+1 and 135/X\n\nsince the rates are the same we can say that the rate for 180/(X+1) is equal to rate for 135/X.  or in equational form it is\n180/(X+1)=135/X.\n\nWe now have a working equation.  If you notice, we are dealing with fractions and to equate it we need to transpose it using cross multiplication.  Cross multiplication is done by pairing the numerator of one side to the denominator of the other side and then proceed to multiplication.\n\nlooking at the working equation, we will pair 180 and X or 180X and another pair which is 135 and (X+1) or 135(X+1)\n\nnow we have a new working equation 135(X+1) = 180X\n\nto simplify 135(X+1) we multiply 135 with X and with 1. 135 and X is equal to 135X and 135 and 1 is equal to 135\n\nnow we have simplified the equation to 135X + 135 = 180X\n\nto get the value of X, we combine similar terms.  they are 135X and 180X.  We combine this by transposition.  Remember that in order not to change the value of the equation we must change the sign of what is moved to another side.  In this case i will move 180X to pair it with 135X and move 135 to the other side. so now we get 135X - 180X = -135\n\nwe simplify the like terms and we get -45X.\n\nnow we have -45X = -135\n\nbut we are only looking for X so we must eliminate -45 from the X.  identify the numerical coefficient of X which is -45 and devide it from the original. what you do to one side, you should also do it to the other.\n\n-45X/-45 = -135/-45 with this equation we can simplify it into\nX=3\n\nnow we have the value for X\n\nso one trip will be 3 and since another is 1 hour longer then it must 3 + 1 = 4\n\nheres the solution\n\n180/(X+1)    =      135/X\n135(X+1)     =      180X\n135X + 135   =    180X\n135X - 180X   =    -135  \n            -45X   =    -135\n         -45X/-45=    -135/-45\n                   X =  3\n          \n             X + 1 =  trip 2\n             3+1 = 4
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1 4 2 5 3\n\nBut it's cyclical, so the sequence is 1 4 2 5 3 1 4 2 5 3 1 4 2 5 3 ... and so on.\nYou can start anywhere in the sequence. The first answer you received is correct, but that person chose to start with 3 instead of 1.
416988
(1) First one is copied wrongly by you                                            (2) Where is the equal to sign?                                                        (3)This equn. also cannot be solved                                              (4)-p/16+5=9  Multiplying all the terms by 16 ,we get -p+80=144   =>-p=144-80=66 Therefore p=66 YOU SHOULD BE MORE CAREFUL BEFORE SENDING YOUR PROBLEMS TO YAHOO ANSWERS.HOWEVER,YOU MAY SEND THE CORRECTED VERSION OF THE PROBLEMS TO ME IMMEDIATELY THROUGH YAHOO ANSWERS.
547434
5 would be (a)\n6 would be (b)\n7 would be (c)\n8 would be (b)