Maratka

coz, it is C

thousands swim up ..., but only few are .... if you would eliminate *but*, you would have got two coordinated clauses thousands smiw up ... and only few .... many species of fish swim up to head water in order to spawn :) what's wrong? Speaking seriously, i will appreciate if someome explain whether "rivers of Maine" can be cosidered a referent for "few of which". IMO, there is no ambiguity as well as author's preferences - thousands of ..., only few of.... Also many of which isn't ambiguous thanks to rules of referent resolutioon. thanks. :rolleyes: :crazy:

spawn by swimming up the rivers of Maine,|but| only a few of which they cant spawn by swimming up... :crazy: but is required in order to introduce clause showing opposition. in other words, b ans c contain different types of clause and convey different meanings.

Meaning of this sentence: a market for bygone ... has grown+2 modifications: 1) a market has grown out of America's fascination with all *things antique; 2) a market has grown, thus bringing* back the chaise lounge, the overstuffed sofa, and the claw-footed bathtub. Author uses inversion in order to emphasize that market has grown out of america's fascination for all things antiques

Agree with Gmat-Help on q96: b only says that p and r have opposite signs. a is sufficient

your answer on q7 is very well-founded. As well as ACB is inscribed angle of the given circumference and points C and O lie on different sides of AB chord then ACB=180-1/2AOB. it is 135 degrees. both statements give us measure of the angle AOB=> we have enough data to calculate angle ACB thank you.

s>1 s+t s+t 1 st1 =>tt sufficient p.s. t0;s0 t>0 st 1) s 2) 0 s>t insufficient

The unknown, If you will count this way then you will get repetitions of the disposal. In other words, fig.1 includes one disposal for each person: for E=A-D, for D:E-C... finally, you will get 4 combinations. in a sense I think that solution of this question is obvious and moreover two alternative approaches was offered. The opinions that has been expressed discord in regarding to how one understand phrase ''the same neighbors". If AC and CA is considered as distinguishable possibilities then the answer is 10. But if AC and CA is considered as indistinguishable possibilities then the answer is 6. IMO, disputing over which variant is preferable is a waste of time.

yep, i'm sorry if iam somehow confused you. 'order important'. i'll review tomorrow it. sorry! i assume that your answer is right, if author consider 123=321, but i don't understand your logic. "pick up numbers" it isn't my approach :) my approach is: 1. find number of combinations when "one person can not have the same neighbors in any two arrangements" 2. reduce number of combination determined in step 1 by number of repetitions

this is why i divided 5!/3! by 2! then :) =5!/(3!*2!)

http://www.www.urch.com/forums/showthread.php?t=11268 i seem to be missing something important http://www.maratka.us/symbols/sits.bmp whats wrong with it? two cases for BAC

had we situation where the order of disposal of 3 persons couldn't influence on the result, then you'd be right. but we have other situation and your statement contradicts conditions of the question: but you state that 123=132=321=312

meanwhile, i have found typo in the original answer. The problem is assumtion that half of odd numbers, excluding "6" are odd. I'll try to explain in terms of your logic. We should count number of integers, excluding 6, to prove this statement. 6 can take position of units, tens and hundred 1) 6**=100 numbers 2) *6*. we have 8 hundreds each has 10 numbers with "6"=80 3)**6. aftter subtraction of 8 tens on the previous step, we have 72 tens (800=80 tens, 80 tens-8tens=72 tens). each of the tens has 1 integer with "6" in units position. => we have 72 integers with "6" in position of unit. --- sum up all 3 numbers: 100+80+72=252. it is here in my previous message: Aftre that i assumed that . But it is incorrect as well as we have: a) 72 even numbers, which have 6 in units. 2) half of remainder is also even (100+80)/2=90 ---- 90+72=162=>252-162=90 odd and,yes, your answer is right. it is typo ... and my mistake :)

I'm not sure whether my answer is correct, but try to clarify my logic. In how many ways we can select three persons fron set of 5 persons? 5!/3!,right? right. then,what does following statement mean for us? one person can not have the same neighbors in any two arrangements It means that we should except combinations in which set of three persons is recurred. for each set of 3 person we have 2! such combinations. It is easy to understand, if one will recognize that we have only 2 persons aside from 3 picked persons. Number of permutations of these 2 persons equals to 2. =>we should shorten initinal set of 5!/3! combinations by 2 =5!(3!*2!)

The best way to keep in mind this conception is to comprehend Euler Venn diagrams. It is very simple math concept Venn Diagrams http://www.maratka.us/symbols/eiler.gif