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depends on ur subject combinationmakaveli said:do i have to study p2 when i prepare for p3?
MarkSkoop1 said:[email protected] first answer would be (12!/4!*2!)=9979200.this is because 4 R are being repeated and 2 E being repeated.
next answer would be (9!/2!)=181440 .....as its said R must be together so taking 4 R as 1,the total arrangement would be 9! and den as 2 E are being repeated in dis case too we will write.......(9!/2!)
the oder last answer would be 6C2.......this is coz when we donot put r in d arrangementwe will hav total letters of 8...and it is even said that the arrangement should contain 2 E..........so now fr formin 4 digits we jus need two more digits form 6 digitS(AFTER DEDUCTING R AND E)hence the answer would be 6C2.........hope u understand.......
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