James ruse 03 trial perm question (1 Viewer)

[Li0n]

All the letters of the word ENCUMBRANCE are arranged in a line.
Find the total number of arrangements, which contain all the vowels in
alphabetical order but separated by at least one consonant.

Yeah its hard

Good luck

 

[ngai]

# ways = 88200 (by inspection)

 

[steverulz55]

88200 ways
i wrote out all the combinations
took me some time

i could scan it for you, but that will be like over 50 pages

 

[Rorix]

i only got 88199 by inspection i must have missed one

 

[Li0n]

:/
the only way is by inspection? no special pattern involved?
if its just inspection then the question isnt as hard as i thought it be.

 

[steverulz55]

awwww Da, better find that missing one!

 

[Archman]

by careful inspection, 88200.

 

[Estel]

# ways = 88200 (by inspection)

88200 ways
i wrote out all the combinations
took me some time

i could scan it for you, but that will be like over 50 pages

i only got 88199 by inspection i must have missed one

by careful inspection, 88200.

88200, by inspection and balance of probabilities

 

[Rorix]

huh how come im getting 88201 now.......


oh wait wooops wrote the one i was missing twice

 

[Trev]

haha ur a funny one rorix

 

[withoutaface]

Tell us how you smug bastards, cause I can't see it

 

[mojako]

:/
the only way is by inspection? no special pattern involved?
if its just inspection then the question isnt as hard as i thought it be.

isnt it harder when u need to do it by inspection???
has anyone seen a pattern now?
Im curious..

 

[Li0n]

isnt it harder when u need to do it by inspection???
has anyone seen a pattern now?
Im curious..

finding a pattern involves a heap of logic. inspection, is just inspection.
which is easy if you know what the hell your doing

i saw a pattern but i didnt know how to implement it kinda

 

[ND]

Here's my solution: (btw, what exactly does "by inspecion" mean?)

Let C denote a consonant and V a vowel. First lay out the Cs:

C C C C C C C

There are now 8 gaps for the Vs, and since there are 4 Vs there are 8C4 ways of distributing the 4 Vs into the 8 gaps. So if all the Cs were identical and all the Vs were identical, there would be 8C4 ways of arranging them. But since there are 7 consonants, with 2 pairs being identical, there are 7!/(2!*2!) ways of arranging them, and since the two E's are indistinguishable, there is 1 way of arranging the vowels. So the answer is:

8C4*7!/(2!*2!)*1 = 88200

 

[mojako]

can you explain why there is 1 way of arranging the vowels? (if we follow what you did for consonants, wouldnt it be 4!/2!?)
EDIT: I forgot they're in alphabetical order.

also, the question says: "which contain all the vowels in alphabetical order but separated by at least one consonant"
What if there are two vowels which are separated by 2 consonants? Maybe you've covered that, I just can't see how
EDIT: Ok, now I see it (thinking while trying to sleep)

Anyway I think the inspection thing is just a joke

Thanks for your help

 

[ngai]

Anyway I think the inspection thing is just a joke

lol u took ur time

 

[withoutaface]

It is by inspection for the sufficiently intelligent (ie top of ruse and imo gold medallist), they just look at the problem and it just jumps out at them

 

[mojako]

It is by inspection for the sufficiently intelligent (ie top of ruse and imo gold medallist), they just look at the problem and it just jumps out at them

hmm is that what inspection means?

I tought it meant.. writing out all possible combinations as steverulz said

 

[withoutaface]

hmm is that what inspection means?

I tought it meant.. writing out all possible combinations as steverulz said

It's just another way of saying: I can do this in my head and cbf writing down the steps, so heres the answer.

Like when you factorise a quartic, I know I hardly ever do that by polynomial division.

 

[mojako]

but... ngai should give the factorials or C notation...
it just takes a few extra characters