Question

Out of 8 consonants and 5 vowels, how many words can be formed so that each word contains two consonants and 3 vowels?

Answer 1

Answer:

Step-by-step explanation:

Given, there are 8 consonants and 4 vowels,

To find words with or without meaning using 3 consonants and 2 vowels :

3 consonants out of 8 can be selected in ⁸C₃ ways = 56

2 vowels out of 4 can be selected in ⁴C₂ ways = 6

So, in total 5 letters of word can be chosen(3 consonants and 2

vowels) in 56*6 = 336 ways

But the chosen set of 5 letter can permute among themselves in

5! ways, hence total number of words would be 336*5!

= 40320

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Answer 2

Given: 8 consonants and 5 vowels

To find: How many words can be formed so that each word contains two consonants and 3 vowels.

Solution:

In order to solve the given question, the concept of combinations can be used. A combination can be calculated as follows.

$^{n}C_{r} = \frac{n!}{(n-r)! r!}$

Here, n is the total number of elements available and r is the number of elements needed in the combination. Since there are 8 consonants and each word must contain 2 consonants, n is 8 and r is 2. Similarly, for vowels, n is 5 and r is 3. Now, the combinations of consonants and vowels must be multiplied by one another because the word contains both consonants and vowels.

$words = ^{8}C_{2} * ^{5}C_{3}$

          $= 28 * 10$

          $= 280$

Therefore, 280 words can be formed so that each word contains two consonants and 3 vowels.