Math Is Fun Forum

yes phrontister, I had hellucination.i wrote down religiously OAEO as mentioned in the problem, noted O should neither follow or precede AE. I thought OA and EO have no such restraint.but if I had written further the sequence AEO I would have realized. Perhaps in the background of my thoughts I felt it is because EAO/OEA  is not barred.

This is how I got my result
N(AE)=no.of arrangements where A and E come together as AE=5!=120
Similarly N(EO=N(OA)=120
N(OAE)=4!=N(AEO)=24
N(AE-OAE-AEO)=Number of arrangements where AE is neither immediately preceded by O nor  immediately followed by O=120-24-24=72
N(OA-OAE)=120-24=96
N(EO-AEO)=120-24=96
So the number of invalid arrangements=96+96+72+24+24=312
No of valid arrangements=720-312=408
Where did I err?

Is there no further discussion on this topic?

f(x) is defined by f(x)=x^3+1 for x ≠ 0 , and at x=0 f(x)=k^2-3k-3.
So f(0) is not to be calculated as 0^3+1 but to be arbitrarily taken as k^2-3k-3.In general it leaves a removable discontinuity at x=0.But for a limited values of k it could become  continuous.this is how I interpret the problem.

k={-1,4} does not come unless L.H.S. is taken as 1.The only thing is Lim x->0 k^2-3k-3 is incorrect as k^2-3k-3 is the value of f(x) at x=0. It is not the limit.The concepts come into play while evaluating ,not the answers.

Seek some clarification.Is( O follow L follow A) i.e. O and A not next to each other allowed?

Hi Bobbym and zetafunc,
I can tell you what Hannibal did. To look  methodical he first took the numerator. It was not his intention to multiply the whole by x-3.Then for each term he found the region where it was positive.Then  he dealt with  the denominator. thereafter he took intersection of all the regions where the terms were found positive.

A rapid method of finding valid intervals( whatever may be the number of factors in numerator and denominator) is count the number of critical points on the right side of an interval. if it is even number( including 0) it is valid interval for the relation >0 type.The reason is simple.All the critical points on the right side of the interval give rise to negative valuefactors.but remember all factors must be of type x- k ,not k-x.

There is a nice analytical approach to the "It could be verse"problem.
If a and  b be the number of two types of cake. 30a+40b=850 or 3a+4b=85
we can consider a packet of 12. Each packet can be distributed as four 3s or three 4s.In 85 there are 12 packets and 1 is left out. The remainder can not be distributed into 3s and/or 4s.So we open one pack ,add to the remainder to make it 13 which is three 3s and one 4.So there are 6 packets left out.We can distribute any number of packets varying from 0 to 6 (in 7 ways) to 3s and the remaing to 4s. thus there are7number of ways.

But a1>=a2.=a3

I think it is quite simple.