Permutations (1 Viewer)

[Aysce]

Q. Find out how many arrangements are possible from the word STUDIO if:
a. The vowels and consonants alternate.
b. The letter O is not first or last.

Thank you very much in advance. I may have more questions to post up later since I honestly suck at these.

 

[LightXT]

Q. Find out how many arrangements are possible from the word STUDIO if:
a. The vowels and consonants alternate.

3 vowels. U I O
3 consonants. S T D

Total number of arrangements = 3*3*2*2*1*1=36

3 possible vowels * 3 possible consonants * 2 possible vowels * 2 possible consonants * 1 possible vowel * 1 possible consonant

36 possible arrangements.

b. The letter O is not first or last.

5 possible letters at the front (excluding O). 4 possible letters at the end (excluding O and first letter). 4! possible combinations in between first and last.
Total arrangements = 5*4*4!
=480 possible arrangements.

I'm not 100% sure if I'm right. It's been a while since I did Maths, and I always lacked confidence for Perms and Combs.

 

[Nooblet94]

I'm pretty sure those are correct (Perms and combs is my weakness)

 

[Aysce]

First question done is incorrect, answer says it is 72. I don't like Perms & Combs :c

 

[Carrotsticks]

First question done is incorrect, answer says it is 72. I don't like Perms & Combs :c

For the first question, the answer is 2 x 3! x 3!

Reason:

- We have 3 vowels and 3 consonants and no repetition. Since we are alternating, this means we can either start with a consonant or a vowel. Hence 2.

- Suppose we have chosen consonants to go first, we can swap them freely amongst each other. Hence 3!

- We now have 3 vowels leftover that we can again swap freely. Hence 3! again.

Therefore, the answer is 2 x 3! x 3!

 

[RealiseNothing]

The first one:

The vowels and consonants alternate, so we can have either:

C V C V C V
or
V C V C V C

V - vowel
C - consonant

Now the first order, we can arrange the vowels in 3! ways, and the consonants in 3! ways. So 3! x 3! ways they can be arranged.

And since the second order is the same as the first but consonants and vowels just swap places, we times the first order by 2.

Hence 3! x 3! x 2 = 6 x 6 x 2 = 72

 

[Carrotsticks]

Oh and to add to that, it's basically Sir's answer, but multiplied by 2 since he forgot to add that it could also begin with a vowel.

 

[Nooblet94]

FFUUU. It's these kind of stupid mistakes that are going to make me end up with an E2 in maths.

 

[RealiseNothing]

For the first question, the answer is 2 x 3! x 3!

Reason:

- We have 3 vowels and 3 consonants and no repetition. Since we are alternating, this means we can either start with a consonant or a vowel. Hence 2.

- Suppose we have chosen consonants to go first, we can swap them freely amongst each other. Hence 3!

- We now have 3 vowels leftover that we can again swap freely. Hence 3! again.

Therefore, the answer is 2 x 3! x 3!

Beat me to it lol.

 

[RealiseNothing]

Also for the second question, here is a quicker method than Sir R's:

The O can not be in the last or first position, so it has to be in one of the middle 4, hence there are 4 sets of arrangements excluding O.

Now the rest of the letters can be arranged in 5! ways, so the answer is 5! x 4 = 480.

 

[LightXT]

>_>