MHB 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 $$(1,3,5,7)$$, and letters $$(2,4,6,8)$$.

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

$$2 \cdot 12 \cdot 12 = 288$$ 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

$$\dbinom{7}{3}\dbinom{5}{0} + \dbinom{7}{2}\dbinom{5}{1} + \dbinom{7}{1}\dbinom{5}{2} = 210$$

 

[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