permutations question (1 Viewer)

[freaking_out]

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.

 

[Rahul]

- _ - _ - _ - _ -

" - " 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:

 

[Fosweb]

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?

 

[Constip8edSkunk]

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

 

[BlackJack]

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

 

[-=«MÄLÅÇhïtÊ»=-]

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.

 

[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?

 

[ND]

Originally posted by -=MLhtʻ=-

bwahaha..actuary..king of probability..

Is probability a major part of actuarial?....... Damn... probability's my worst topic....

 

[freaking_out]

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.

 

[-=«MÄLÅÇhïtÊ»=-]

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

dun worry too much
im not anything special in probability either
but they'll go thru the diff types of combs and perms and everything becomes clear.

 

[ND]

That's good to hear.

 

[ND]

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?

The 6C4/2! is like the possible number of ways it can be 'structured', and the 720 is the number of ways the letters can be arranged (in their cons/vowel respective positions).

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