oh damn..........this is such a bloody hard q......i remember givin this to ma teacher and he couldnt solve it..........
there are two different cases you need to consider......the first is that he selects 3 sweets all of different kinds......u knw that he can choose any 3 flavours frm the 6 available.....the number of selections is 6C3=20
the second case is that 2 of the sweets are the same and the other one is of a different flavour. when he chooses the first sweet, he has a choice between 6 different flavours. when he chooses the second sweet he still has a choice between 6 flavours since he can select another of that
flavour. but when he selects the third sweet he has a choice of 5 flavours since two sweets have already been taken form the sixth flavour and he
cannot take a third. the number of selections is 6*5=30
so the total is 30+20=50
OMG LOL i asked my teacher as well and i think he explained it 4 times but i still had NO IDEA what he was talking about.
i did get it for a second but then i forgot D:
ANYWAYS;
'the second case is that 2 of the sweets are the same and the other one is of a different flavour. when he chooses the first sweet, he has a choice between
6 different flavours. when he chooses the second sweet
he still has a choice between 6 flavours since he can select another of that
flavour. but when he selects the third sweet he has a choice of
5 flavours since two sweets have already been taken form the sixth flavour and he
cannot take a third. the number of selections is 6*5=30'
ummm i get the first part but for the second case, why is it not 6*6*5 if he has a chance of choosing 6 diff flavours the first time, 6 the next, & 5 on his third choice?
thankyouuuus.
