How many different ways can the letters of word education be arranged so that no two vowels are together?

i think this question my teacher gave me is from a past paper:

find the number of ways in which the letters of the word "EXTENSION" can be arranged in a straight line so that no two vowels are next to each other.

- _ - _ - _ - _ - " - " represents the position of the consonants. their possibilites are 5!/2! " _ " possible positions of the vowels. 4!/2! (5!/2!) x (4!/2!) = 720

i dont know if thats right. :mad1:

i think its a little more complicated than that... you are missing a few arrangements where they still arent next to each other... like say: _---_-_-_ (3 in this form - with three consonants together) and: -_--_-_-_ (5 in this form - with two consonants together) then the ones you have: -_-_-_-_- (9 in this form - with no consonants together) then work out arrangements... is this really a 3U question? it seems hard...???

so how do you do the rest?

wouldnt it b easier subtracting the number of arrangements when 2 vowels r next 2 eachother from total arrangements

Think of it this way... (Pidgeon hole style) you have 5 consonants making 6 places (holes) where you can only placee one vowel. You have 4 vowels to place among those 6 positions (!). 0|0|0|0|0|0 Therefore, 6C4 is the combinations. (15 ways.) Now, the letters themselves: 720 like Rahul said. So 15*720 = 10,800. note: These are the 15, if you want to check. 0|0|0|0|| 0|0|0||0| 0|0|0|||0 0|0||0|0| 0|0||0||0 0|0|||0|0 0||0|0|0| 0||0|0||0 0||0||0|0 0|||0|0|0 |0|0|0|0| |0|0|0||0 |0|0||0|0 |0||0|0|0

||0|0|0|0

there are 4 vowels and 5 cons since vowels cant b put together we can only put 1 vowel between 2 cons so if u can visualise the situation, the vowels can only go in (6P4)/2! (divide by 2! because there are 2 E's) places and the cons can be arranged in 5!/2! (divide by 2! because there are 2 Ns) ways so answer is (6P4)/2! * 5!/2! = 10800 bwahaha..actuary..king of probability..

edit: this is the type of ques that you select first, then permutate. It's made abit more tricky den standard questions coz there are repititions.

i like ur way blackjack....jus one question, why do you multiply the number of possible arrangements by 6C4/2!? i know 6C4/2! is the number of possible arrangements of the consonants around the vowels, but what is the principal behind it?

Originally posted by -=MLhtʻ=-
bwahaha..actuary..king of probability..

hey rahul, take this question to arnold , coz i found out that its from a past catholic 3u paper, and the answer has 720 in it....so theres probably a mistake in the answers (these were official answers as well).

i'am a bit confused.... but i will post up the solutions to this question soon.

Originally posted by ND
Is probability a major part of actuarial?....... Damn... probability's my worst topic....

but they'll go thru the diff types of combs and perms and everything becomes clear.

Originally posted by Rahul
i like ur way blackjack....jus one question, why do you multiply the number of possible arrangements by 6C4/2!? i know 6C4/2! is the number of possible arrangements of the consonants around the vowels, but what is the principal behind it?

I don't know if that made sense...