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So the limit N goes to infinity is here.
Читать полностью…now I understand how to apporach the question
Читать полностью…This gives the missing 6! @Peacefulmaninhouse
Читать полностью…I am not sure if I miss one configuration. I am exactly 6! short of 7200.
Читать полностью…ouh I see.So it will be like this?!
Читать полностью…But possibly can occupy one end.
Читать полностью…The second term corresponds to the situation 5 non-E letters are together. (I don't have pen &paper right now)
Читать полностью…because when I see the plus sign between terms I definitely we need to approach by cases but which one to start first?
Читать полностью…ouh wait how does one tackle the question?
Читать полностью…for part (ii).I thought about the idea of fixing the 3 Es and solving but when I checked the answer.I might have undercounted but I dont know what the answer key working explains.Can someone help.
Читать полностью…@qqqqqwwwwweeerrrrt has nailed it. I add some comments below.
Читать полностью…Anyone can help pls? Appreciate loads ya! Tksss
Читать полностью…hi can anyone help to check if im doing part ii correctly?
Читать полностью…ouh I see.Thanks guys for helping me.
Читать полностью…My apologies. This is exactly the second term describes. My previous guess for the second term is wrong. I will try to this problem later tonight.
Читать полностью…This case has been considered in the third term. The other two terms are for situation where E's don't occupy both ends.
Читать полностью…The first term in the key corresponds to the situation where 4 non-E letters are together, and there are 3 arrangements arrangements for sets of this four letters set, and one letter set and another one letter set.
Читать полностью…is it we have to count how many ways there are to arrange the three Es in that sense
Читать полностью…Quick reply. The last term corresponds to the situation where two E's are on both ends and the third E have three options for it's place.
Читать полностью…For thoae who has done calculus: Setting the derivative of the given expression to zero gives the expression on the left box of this picture.
Geometrical interpretation is the triangles APC and BPD are similar. This is possible only when P becomes O for a suitable x.
The segments AP+PB will be minimum when P shifts to O. When this happen AP+PB will be the segment AB, whose length is sqrt(12^2+5^2)=13.
Читать полностью…the qn uses x and 12-x, so a reasonable starting point would be to draw 2 triangles with base x and 12-x together, and the rest of the proof should follow naturally
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