Combinatorics: How many Words with 2 Different Vowels and 2 Different Consonants can be Formed from Alphabet?

Combinatorics is a branch of mathematics that focuses on studying the selection, arrangement, and operation of countable discrete structures. This is essential in computer science because it can be used to solve problems regarding statistics and probability. Moreover, combinatorics also plays a big role in the optimization of various applications. Combination and Permutation are two terms that are often used to solve problems in combinatorics.

Combination and Permutation

Combination in simple terms is a method for selecting elements from a collection where the order of selection does not matter.

Whereas, permutation is the arrangement of elements from a collection. The elements of permutation should be arranged in a particular order.

Example:

Suppose we have 3 variables named x, y, and z.

Find the combination and the permutation of two variables from the given variables

Combination

Combination: xy, yz, and zx

Permutation: xy, yx, yz, zy, zx, and xz

From the example above, we can see that in the permutation both the xy and yx are written because the order of arrangement matters. However, in the combination, only one of them is written because both are the same things since the order of arrangement does not matter.

Formula:

Combination = C(n,r) = n! / (r! × (n – r)!)

Permutation = P(n, r)= n! / (n – r)!

Note:

n: It is the size of the set

r: It is the size of the subset of n

How many words with 2 different vowels and 2 different consonants can be formed from the alphabet?

To solve this problem, we have to know that the alphabet contains 26 letters including 5 vowels and 21 consonants.

Step 1:

Ways of selecting 2 different vowels out of 5 vowels

C(5, 2) = 5! / (2! × (5 – 2)!) = 5! / (2! × 3!) = 10

Step 2:

Ways of selecting 2 different consonants out of 21 consonants

C(21, 3) = 21! / (2! × (21 – 2)!) = 21! / (3! × 19!) = 210

Hence, combinations of 2 different vowels and 3 different consonants is 10 × 210 = 2100

Step 3:

Ways arranging or forming 2 vowels and 2 consonants out of 4 letters

P(4, 4) = 4! / (4 – 4)! = 4! / 0! = 4! = 24

Therefore, the total number of ways is 2100 × 24 = 54000

∴ The answer: 54000 words

FAQ on Combinatorics

Q.1: What is the real-life example of combinations?

Answer: 

Picking three finalists of a competition, picking colored balls from a box, etc.

Q.2: What is the real-life example of permutations?

Answer: 

Sequence of numbers for a lock, car plate numbers, etc.

Q.3: What is the main difference between permutation and combination?

Answer: 

In simple word, the order of arrangement matters in case of combination, but it does not matter in permutation.

Q.4: Is there any alternative to find the combination other than using the formula?

Answer: 

Yes, there is. Pascal’s Triangle can also be used to find combination.