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(u will never get asked a card question in the HSC people!) 2) How many different sums of money can be made up from six $20 notes, two $5 notes, four $2 coins, seven $1 coins, three 50c coins, and one 20c coin ?? the answer is not 588 peoples, this is how the book did it which i cannot...

ohhh, yeh... sorta :giddy:

i like spice girl's answer for the first one and blackjacks answer for the second one is heaps good, nice logic there, the way they did it in the textbook was long and stoopid thanx people :wave:

i do 4 maths and the maths in physics and chem is non existence, for real both subjects are shitty memorising and rote learning tables. doing all those subjects if i were u i'd drop one of the 4 unit either maths or english cause man u will be non stop working and still going shit with that...

1) How many different six letter words can be formed from the letters of the word AUSTRALIA ?? 2) How many different sums of money can be made up from six $20 notes, two $5 notes, four $2 coins, seven $1 coins, three 50c coins, and one 20c coin ??

Re: Re: Re: combination? how has it?? please explain so what does that mean in terms of the 3 letter word> and why divide by 2!

ok, can someone tell me what i'm doing wrong please!?? equation 5 vowels 3 consonants 1vowel * 1consonat * any other letter 5*3*6 = 90 way now these 3 letters can be arranged 3*2*1 ways = 6 so 90*6 = 540 ways

yehha i got that as my answer!, sweet

The deck of a ship was 2.4m below the level of a wharf at low tide and 0.6m above wharf level at high tide. Low tide was at 8.30a.m and high tide at 2.35 p.m. Find when the deck was level with the wharf, if the motion of the tide was simple harmonic?? [i know how to do it, i just dont have...

i thought AM>=GM?? where does HM fit in..

cool ive got nothing, how do u do it??

yeh true thats right good stuff :guitar:

a circle geom question seems to be a favourite reduction formulae induction ( which i hate) graphing absolute value, derivate, inverse and that sorta stuff of a certain graph

ohh wait, yeh mines wrong, i did when it was AT MOST 2 go out the same exit which is 1- (1/5*3 + 4c3 * 1/5^3 * 4/5^2) yours is right

hold up ahh i did it this arvo and i forgot now its something like that

for the second one the answer is 1 - ( 1/5^3 + 4c3*1/5^2 *4/5^3)

no wait laz your right, the fact that its only 1/5^3 for the first one because it is dependant on whatever exit the first one goes to good stuff

wrong !! :mad1:

wrong !! :mad:

four rats are in a maze, and there a five exits. each rat is equally likely to exit any maze, so the probability of a rat exiting by a certain exit is 1/5. 1) what is the probabilty of all four rats leaving the maze by the same exit? 2)what is the probability of at least 2 rats leaving by...