combs - ALGEBRAIC vowels arrangement 01 HSC q2c (1 Viewer)

[gamja]

How many arrangements of the letters in the word ALGEBRAIC are possible if the vowels must occupy the 2nd, 3rd, 5th and 8th positions?

ALGEBRAIC is a 9-letter word
Vowels are AAEI, consonants are LGBRC
I would solve this using cases:

1. AA in 2rd, 3rd - e and i in 5th, 8th, alongisde 5 consonants arranged = 2!x5! = 240
2. AZ in 2nd, 3rd - each e and i being next to a, alongside 5 consonants arranged = 2!x2!x2x5! = 960
3. ZZ in 2nd, 3rd - e and i arranged in 2nd, 3rd, 5 consonants arranged = 2!x5! = 240

total 1440 arrangements

Is there a faster way to solve this without using so many cases? I usually find myself in harder perms and combs to just grind through cases forever - i need to practise using faster ways.

Thanks in advance!

 

[011235]

There are 4 places to place the vowels. Ordering those: 4!/2!

There are 5 places to place the consonants. Ordering those: 5!

Hence there are 4!*5!/2! = 1440 ways.

What drew you to use cases for this question?

 

[gamja]

There are 4 places to place the vowels. Ordering those: 4!/2!

There are 5 places to place the consonants. Ordering those: 5!

Hence there are 4!*5!/2! = 1440 ways.

What drew you to use cases for this question?

that is a much faster way - i was drawn to use cases because i got baited by the separate positions for the vowels when actually it doesn't matter for vowel arrangement

 

[011235]

yep, as a general rule, you only need use cases if something is dependent on something else. for example, if placing the consonants affected placing of vowels, or smth like that

 

[5uckerberg]

There are 4 places to place the vowels. Ordering those: 4!/2!

There are 5 places to place the consonants. Ordering those: 5!

Hence there are 4!*5!/2! = 1440 ways.

What drew you to use cases for this question?

Notice that in the division of is because you have 2 A's as you assume the 2 A's are going to be the same.