# In how many ways a party of 16 people be conveyed in 2 vehicles of which one will hold not more than 8 and other not more than 10

$Thereare16peopleinthepartywhohavetobeconveyedin2vehicles,ofwhich,onewillnotholdmorethan8people,whiletheotherwillnotholdmorethan10people.\phantom{\rule{0ex}{0ex}}Letthenumberofpeoplegoinginthefirstandsecondvehiclesbexandyrespectively.\phantom{\rule{0ex}{0ex}}Then,x+y=16;x\le 8andy\le 10.\phantom{\rule{0ex}{0ex}}Now,theaboveequationsaresatisfiedwhen(x,y)=(6,10);(7,9);(8,8).Thus,thereareatotalof3waysofconveyingthe16peopleoftheparty.\phantom{\rule{0ex}{0ex}}Note:Ithasbeenassumedthateachvehiclemakesoneandonlyonetripforconveyingthe16peopleoftheparty.$

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