One way u were thinking was: we assign 5 periods to the 5 subjects in 6P5 ways : 720 ways the remaining period can get ne one subject in 5 ways . so total no. of ways = 720 * 5 = 3600 ways Second way was: Each of 5 subjects has to be given atleast 1 period , so no: of ways 5 subjects fit to 5 peiods is 5P5 =120 . Then the remanining period can be filled in 5 ways ( considering that the ramining period has to have 1 subject and is not free) . That makes it 120 *5 =600 . This whole thing can be done in 6 ways , i.e either of the six periods could be the one which we are calling the remaining period. Hence 6*600 =3600 . Had it been possible that the remaining period is vacant , then the answer would have been 120*6*6 =4320 Slackdemic: the 6 periods where 2 r idenctical can be taught in 6!/2! = 360 ways there r 5 subjects and any of them can have 2 periods, so the ans is 360*5 = 1800 coz your method is not taking into account dat : suppose subject maths has 2 periods in a day at 1st and 2nd periods........ then ur method counts this twice[ie 1st & 2nd period and 2nd & 1st period as separate]........coz u r treating maths at 1st and 2nd perids as different subjects.........which is wrong ......so ur answer shud be divided by 2........ This might be ur solution.