# Vowel-Consonant Arrangements and Non-Adjacent Vowels

• MHB
• DaalChawal
In summary, there are 720 ways to arrange 4 vowels and 4 consonants in 9 positions such that the vowels are never adjacent to each other. This is calculated by first arranging the 4 consonants in their designated places and then arranging the 4 vowels in the 5 possible gaps between the consonants. This method accounts for all possible 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

Question itself and options 1 and 3.

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.

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

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

Thank you so much!

romsek said:
So gathering all this up with have

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)

$a - b$

DaalChawal said:
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.

Thanks, I understood it now.

@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

## 1. What is the difference between permutation and combination?

Permutation refers to the arrangement of a set of objects in a specific order, while combination refers to the selection of a subset of objects from a larger set without regard to order.

## 2. How do I calculate the number of permutations or combinations?

The number of permutations can be calculated using the formula n!/(n-r)! where n is the total number of objects and r is the number of objects being arranged. The number of combinations can be calculated using the formula n!/r!(n-r)!, where n is the total number of objects and r is the number of objects being selected.

## 3. Can repetition be allowed in permutations and combinations?

Yes, repetition can be allowed in permutations and combinations. In permutations, repetition refers to the possibility of an object being used more than once in a specific arrangement. In combinations, repetition refers to the possibility of an object being selected more than once in a subset.

## 4. How are permutations and combinations used in real life?

Permutations and combinations are used in various fields such as mathematics, statistics, computer science, and engineering. They are used to solve problems involving probability, counting, and optimization. For example, they can be used to calculate the number of possible outcomes in a game of chance or to determine the number of possible seating arrangements at a dinner party.

## 5. What are some common misconceptions about permutations and combinations?

One common misconception is that the order of objects does not matter in combinations. In reality, the order of objects does matter in combinations, as changing the order would result in a different combination. Another misconception is that permutations and combinations are only used in mathematics. In fact, they are used in various fields and have practical applications in real life.

• General Math
Replies
1
Views
996
• General Math
Replies
1
Views
2K
• General Math
Replies
4
Views
1K
• General Math
Replies
3
Views
2K
• General Math
Replies
2
Views
2K
• General Math
Replies
1
Views
799
• Precalculus Mathematics Homework Help
Replies
5
Views
1K
• General Math
Replies
2
Views
1K
• Math Proof Training and Practice
Replies
23
Views
966
• General Math
Replies
2
Views
2K