sgEdu 🇸🇬 MATH & Math'l Sciences

27 Jan 2021 13:48

Thank u so much Miss Lee!

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 16:08

So the limit N goes to infinity is here.

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 01:40

now I understand how to apporach the question

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 01:39

This gives the missing 6! @Peacefulmaninhouse

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 15:33

I am not sure if I miss one configuration. I am exactly 6! short of 7200.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 15:31

I get 6480. @Peacefulmaninhouse

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:20

ouh I see.So it will be like this?!

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:16

But possibly can occupy one end.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:11

do you mean like this?

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:59

The second term corresponds to the situation 5 non-E letters are together. (I don't have pen &paper right now)

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:53

because when I see the plus sign between terms I definitely we need to approach by cases but which one to start first?

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:51

ouh wait how does one tackle the question?

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:23

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

23 Jan 2021 17:11

😔, please correct to 5.

sgEdu 🇸🇬 MATH & Math'l Sciences

23 Jan 2021 17:05

@qqqqqwwwwweeerrrrt has nailed it. I add some comments below.

sgEdu 🇸🇬 MATH & Math'l Sciences

27 Jan 2021 13:43

Anyone can help pls? Appreciate loads ya! Tksss

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 15:43

hi can anyone help to check if im doing part ii correctly?

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 01:40

ouh I see.Thanks guys for helping me.

sgEdu 🇸🇬 MATH & Math'l Sciences

26 Jan 2021 00:18

Still got _E_ _E_ _E_

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 15:32

eh then why the answer is 7200?

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:26

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:17

In the first two terms.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:16

This case has been considered in the third term. The other two terms are for situation where E's don't occupy both ends.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 13:04

I mean 3! arrangements

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:57

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:52

is it we have to count how many ways there are to arrange the three Es in that sense

sgEdu 🇸🇬 MATH & Math'l Sciences

25 Jan 2021 12:50

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

23 Jan 2021 18:21

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

23 Jan 2021 17:10

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.

sgEdu 🇸🇬 MATH & Math'l Sciences

23 Jan 2021 11:27

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