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March 29, 2024, 12:37:26 am

Author Topic: 3U Maths Question Thread  (Read 1230256 times)  Share 

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RuiAce

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Re: 3U Maths Question Thread
« Reply #135 on: May 10, 2016, 09:23:59 am »
+1
help with this ques thanks
let n be a positive even integer
expand and simplify
(a+b)^n + (a-b)^n


what i got was
2(nC0)a^n + 2(nC2)(a^(n-2))b^2 + 2(nC2)(a^(n-4))(b^4) ....+ 2(nCn)(b^n)

This looks right

IkeaandOfficeworks

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Re: 3U Maths Question Thread
« Reply #136 on: May 10, 2016, 11:35:04 am »
0
Hi guys! I need help with this question:

 tan^-1 (3/4) = 2tan^-1 (1/3)

Thank you!

RuiAce

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Re: 3U Maths Question Thread
« Reply #137 on: May 10, 2016, 12:13:25 pm »
+2
Hi guys! I need help with this question:

 tan^-1 (3/4) = 2tan^-1 (1/3)

Thank you!


katherine123

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Re: 3U Maths Question Thread
« Reply #138 on: May 14, 2016, 08:57:25 pm »
0
how to find coefficient of x^8  in the expansion of (x^2-2x+3)^5 

RuiAce

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Re: 3U Maths Question Thread
« Reply #139 on: May 14, 2016, 09:37:41 pm »
+3
how to find coefficient of x^8  in the expansion of (x^2-2x+3)^5






amandali

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Re: 3U Maths Question Thread
« Reply #140 on: May 17, 2016, 05:45:44 pm »
+1



ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together

im not sure what i did wrong but ans is  2/7

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #141 on: May 17, 2016, 10:47:55 pm »
+2
(Image removed from quote.)
ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together
im not sure what i did wrong but ans is  2/7

Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently:

Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply:



Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives:



Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore:



I hope this helps! That was a tricky one, stumped me for a bit there!  ;D

jakesilove

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Re: 3U Maths Question Thread
« Reply #142 on: May 17, 2016, 11:30:47 pm »
+2
Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently:

Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply:



Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives:



Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore:



I hope this helps! That was a tricky one, stumped me for a bit there!  ;D

This is the first time I've had a problem with factorials being donated as '!'. Reading through your (well written) answer, I kept seeing a super enthusiastic Jamon explaining how probability works. 'Now, we just sub this in! Then, divide by this! Isn't maths just so very fun!'.
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Re: 3U Maths Question Thread
« Reply #143 on: May 18, 2016, 11:36:57 am »
+1
This is the first time I've had a problem with factorials being donated as '!'. Reading through your (well written) answer, I kept seeing a super enthusiastic Jamon explaining how probability works. 'Now, we just sub this in! Then, divide by this! Isn't maths just so very fun!'.

Ahaha, but maths is so very fun, right?  :o

RuiAce

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Re: 3U Maths Question Thread
« Reply #144 on: May 18, 2016, 05:40:42 pm »
+1
Confession: I purposely didn't answer that question because I don't want to associate with perms and combs again until next semester when I'm doing it in discrete maths.

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #145 on: May 18, 2016, 10:24:03 pm »
+1
Confession: I purposely didn't answer that question because I don't want to associate with perms and combs again until next semester when I'm doing it in discrete maths.

Ahaha! Well if you handle the probability questions next semester I'll handle the integrals  ;)

RuiAce

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Re: 3U Maths Question Thread
« Reply #146 on: May 19, 2016, 01:01:09 pm »
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Ahaha! Well if you handle the probability questions next semester I'll handle the integrals  ;)

I can handle probability :P just want to avoid perms and combs

amandali

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Re: 3U Maths Question Thread
« Reply #147 on: May 20, 2016, 08:22:34 pm »
0



need help with this ques thanks

RuiAce

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Re: 3U Maths Question Thread
« Reply #148 on: May 20, 2016, 09:06:25 pm »
+1
(Image removed from quote.)


need help with this ques thanks







Worthwhile mention: The harder binomial coefficient proofs always include at least one of the following:
Symmetry property of binomial coefficient: nCk=nCn-k
Pascal's identity:n+1Ck+1=nCk+nCk+1

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #149 on: May 20, 2016, 11:03:37 pm »
+1






Worthwhile mention: The harder binomial coefficient proofs always include at least one of the following:
Symmetry property of binomial coefficient: nCk=nCn-k
Pascal's identity:n+1Ck+1=nCk+nCk+1

I remember the first time I tried this proof in my HSC I tried to do it with the full factorial expansions of each term, ever try it that way? I mean, not that you ever would, my teacher quickly steered me right  ;D