Vowel-Consonant Arrangements and Non-Adjacent Vowels

[DaalChawal]

Question itself and options 1 and 3.

 

[romsek]

There are 2 sets of letter positions vowels can be placed in such that they are never adjacent.
These are letters \(\displaystyle (1,3,5,7)\), and letters \(\displaystyle (2,4,6,8)\).

Within these slots there would be \(\displaystyle 4!=24\) arrangements of the 4 vowels however there are 2 of the letter "A".
Thus we have \(\displaystyle \dfrac{4!}{2!} = 12\) different arrangements of the 4 vowels.

Now we have 4 consonants to arrange, two of which are the letter "R".

This is essentially the same situation as with the vowels. Arranging 4 letters, 1 of which is duplicated, into 4 positions.
So as before there are 12 ways to arrange these consonants.

So gathering all this up with have

\(\displaystyle 2 \cdot 12 \cdot 12 = 288\) distinguishable arrangements with no vowels adjacent to one another.

 

[romsek]

2) is simple. Just look at how many valid digit choices you have for each digit of a 3 digit number.
Note that 0 is not included as a possible digit and so we can't use the number 100 anyway so we don't have
to worry that we don't use every 3 digit number.

3) Use stars and bars

4) Is a bit trickier. A triangle can have at most 2 colinear points. Consider two groups.
The group of 5 colinear points, and the the group of the other 7 points.
There are 3 ways to form these triangles.

i) use 3 points from the set of 7
ii) use 2 points from the set of 7 and 1 point from the set of 5
iii) use 2 points from the set of 5 and 1 point from the set of 7

We can read off the number of arrangements of these simply enough to get a total of

\(\displaystyle \dbinom{7}{3}\dbinom{5}{0} + \dbinom{7}{2}\dbinom{5}{1} + \dbinom{7}{1}\dbinom{5}{2} = 210\)

 

[DaalChawal]

Thank you so much!

So gathering all this up with have

Instead of this method If I solve like this I think I'm doing some error. Can you please help

Total permutation of all 8 letters = $\frac{8!}{2! 2!}$ = (a)

and now let's assume all the vowels are together so ways will be = $\frac{4!}{2!}$ = (b)

Now the required answer is
$a - b$

 

[romsek]

and now let's assume all the vowels are together so ways will be = $\frac{4!}{2!}$ = (b)

You are omitting all the cases where two vowels are adjacent but the others are not, or where two pairs of vowels are adjacent but separated by a consonant.

 

[DaalChawal]

Thanks, I understood it now.

 

[DaalChawal]

@romsek
What is wrong in this method?
_ _ _ _ _ _ _ _ _ (Example --> _R_R_I_M_ There are 5 gaps)
Number of vowels to be never adjacent = (Number of ways in which the 4 consonants can be arranged in their designated places) × (Number of ways in which the 4 vowels can be arranged in the 5 possible places) = $\frac{4!}{2! }$ . $5 \choose 4$ . $\frac{4!}{2! }$ = 720