Why do i suck at probability / permutations & combos? (1 Viewer)

[Giant Lobster]

I hate the world... life is so cruel...

how many ways can REMEMBER be arranged if the Es are not to be next to ea other?

 

[Rahul]

_ _ _ _ _ _ _ _

4 x 5!/2!x2!
120

is that right?

4 -there are 4 ways that the E's can be placed in the blanks above. [HINT: actually draw it out the blanks and work on it as u go]

5! -there are 5 letters that can be arranged in the remaining spaces

2! -due to the repetition of R and M.

am i right?

 

[Constip8edSkunk]

hmm i got 600...

 

[Rahul]

Originally posted by Constip8edSkunk
hmm i got 600...

i must be wrong, what was ur method?

edit: i didnt muliply it by the number of times the groups could be ordered....which is 5.
5 x 120 = 600

 

[Constip8edSkunk]

quite messy actually, prolly there should b a clearer way...

(8!/24) - ((5*6)*5!/4) - (6!/4)
total words - 2Es 2gether - 3Es 2gethr

edit: come2 think of it u can prolly just subtract any 2 Es together, without separating it out, actually subtracting isnt that good an idea.... meh brain too dead after summarising chem arghh...

 

[Affinity]

600

arrange R,R,M,M and B -> 30 ways (5!/(2!2!))
but then you should multiply by 6C3 since there are 6 'gaps' (not 4 since the 2 ends are also gaps!)

 

[Giant Lobster]

so complicated... and this qu was from a yr 11 past paper. ugh i hate permus & combos. thanks anyways ppl.

 

[Giant Lobster]

Aiight two more questions. if stuff is replaced then u cant use combo and permu to do it right?

e.g. a combination safe, 4 digits. obviously there are 10000 different permutations (10 numbers per digit ^ 4) but how wud u express that in maths?

and how do u do a question like this:

AAAABBC pick 4 how many combos?

 

[deyveed]

Hmm... maybe its 7C4/4!2!
I dunno. Just having a go

 

[ND]

Originally posted by Giant Lobster
Aiight two more questions. if stuff is replaced then u cant use combo and permu to do it right?

e.g. a combination safe, 4 digits. obviously there are 10000 different permutations (10 numbers per digit ^ 4) but how wud u express that in maths?

and how do u do a question like this:

AAAABBC pick 4 how many combos?

I don't understand what you mean for the first one. With the second, take cases, i.e:

All same letter, 3 same letter 1 different, 2 same 2 same, 2 same 2 different.

Work out how many possibilities there are for each case, and add them together.

 

[BlackJack]

First, just 10^4. When letters are replaced, you generally have an exponential instead of a factorial.

Second, yes, by summing the parts.

 

[Giant Lobster]

ahh i see
so when stuff is replaced it has nufin to do with permus and combos

and whole idea of adding the possibilities of cases to me seems really tedious / unreliable / room for error

wat if ur givin a really really gay question with lots of cases. ull be hardpressed not making a few arithmetic errors on the way. but then i guess sometimes that is the only way..

 

[kouri]

the probability of

- being crap at probability and perm/com = 9/10
hmm...

theres still 10%.... keep your hopes up guys

 

[...]

for some reason, i love this topic...

 

[shadowRRL]

Originally posted by Giant Lobster
Aiight two more questions. if stuff is replaced then u cant use combo and permu to do it right?

e.g. a combination safe, 4 digits. obviously there are 10000 different permutations (10 numbers per digit ^ 4) but how wud u express that in maths?

and how do u do a question like this:

AAAABBC pick 4 how many combos?

I would say 7C4/4!2!

 

[freaking_out]

hey Giant Lobster, your sig. is cool....which website did u get it from?

 

[jm1234567890]

permutaions and combinations will never be a large part of HSC.

I suck at them and i got 97 in HSC.

I guess I was luck since only 4 easy marks on them in 2002

 

[freaking_out]

Originally posted by jm1234567890
...i got 97 in HSC....

wow, i wish i can get such a result.