How many words of 4 consonants and 3 vowels can be formed out of 8 consonants and 5 vowels?

There are 3 consonants and 3 vowels.
So, 4 consonants and 2 vowels can be selected in 8C4 × 3C2 ways.
Now, 8C4 × 3C2 

= `[[8 xx 7 xx 6 xx 5]]/[[4 xx 3 xx 2]] xx [[3 xx 2 xx 1]]/[[2 xx 1]`

= 70 × 3
= 210
Thus, there are 210 groups consisting of 4 consonants and 2 vowels.
We need to form different words from these 210 groups.
Now, each group has 6 letters.
These 6 letters can be arranged amongst themselves m 6! Ways.
∴ The number of required words
= [210] × 6!
= [210] × 720
= 151200

4 consonants can be selected from 8 consonants in 8C4 ways and 2 vowels can be selected from 3 vowels in 3C2 ways.

∴ the number of words with 4 consonants and 2 vowels = 8C4 × 3C2

= `[8!]/[4!4!] xx [3!]/[2!1!]`

= `[8 xx 7 xx 6 xx 5]/[4 xx 3 xx 2 xx 1] xx [3 xx 2!]/[2!]`

= 70 × 3

= 210

Now each of these words contains 6 letters which can be arranged in 6P6 = 6! ways.

∴ the total number of words that can be formed with 4 consonants and 2 vowels

= 210 × 6!

= 210 × 6 × 5 × 4 × 3 × 2 × 1

= 151200.

Hint: To solve the question you need to know the knowledge of combination and permutation. We need to calculate the combination for the selection of the consonants and vowels, using the formula which is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left[ n-r \right]!}$ . The next step will be the arrangement of the letters which will be done using permutation, using formula ${}^{n}{{P}_{r}}=\dfrac{n!}{\left[ n-r \right]!}$.

Complete step-by-step solution:
The question asks us to find the number of distinct words that can be formed using $4$ constant and $2$ vowels if we are given with $8$ consonants and $3$ vowels.
The first step to solve the question will be to find the number of groups that can be formed using $4$ constant and $2$ vowels when we are given with $8$ consonants and $3$ vowels. Total number of groups formed will be calculated using combination, so the number of groups that can be formed are:
$\Rightarrow {}^{8}{{C}_{4}}\times {}^{3}{{C}_{2}}$
On calculating the combination given above we get:
$= \dfrac{8!}{4!\left[ 8-4 \right]!}\times \dfrac{3!}{2!\left[ 3-2 \right]!}$
$= \dfrac{8!}{4!\left[ 4 \right]!}\times \dfrac{3!}{2!\left[ 1 \right]!}$
Now we will be solving for the factorial of the numbers present in the expression above. The factorial of the number $n$ is represented as $n!$ which is the product of all the numbers from $1$ to $n$ . The formula which we will use here will be $n!=n\left[ n-1 \right]!$ . So on applying the same we get:
$= \dfrac{8\times 7\times 6\times 5\times 4!}{[4\times 3\times 2\times 1]4!}\times \dfrac{3\times 2!}{2!\left[ 1 \right]!}$
Cancelling out the terms common in numerator and in the denominator we get:
$= \dfrac{7\times 6\times 5}{1}\times \dfrac{1}{1}$
On final calculation we get:
$=210$
The number of groups of letters found to be $210$ . To find the total number of words formed we will multiply $210$ to the factorial number of letters as different arrangements of the letter will form different words. Here we will be using the concept of permutation as we are required to arrange the letters. So on calculating we get:
$= 210\times {}^{6}{{P}_{6}}$
$= 210\times 6!$
On calculating we get:
$= 210\times 6\times 5\times 4\times 3\times 2\times 1$
$= 151200$
$\therefore $ Total distinct words that can be formed using$4$ constant and $2$ vowels from has $8$ consonants and $3$ vowels is $151200$.
Note: For the selection purpose ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left[ n-r \right]!}$ is used, which actually determine the number of possible arrangements. Factorial of a number is the product of all the numbers from one to the number itself, which means $n!=n[n-1][n-2]....2.1$. The formula which has played an important role in the above solution and is helpful is to expand $n!$ as the product of numbers and factorials. So $n!$ could be written in many ways like $n!=n[n-1]!$ , $n!=n[n-1][n-2]!$ and so on, as per the demand of the problem.

Simplify problem.

3 consonants BCD.
3 vowels AEI.

Choose 2 consonants, no repetition.
Choose 2 vowels, no repetition.

Form different words from the chosen consonants and vowels.

Choose 2 consonants, 3 ways - BC, BD, CD.
Choose 2 vowels, 3 ways - AE, AI, EI.

Different groups with 2 consonants and 2 vowels, 3x3 = 9.

BCAE, BCAI, BCEI, BDAI, BDAI, BDEI, CDAE, CDAI, CDEI.

Different words from 4 character set such as BCAE, 4! = 4x3x2x1 = 24.

BCAE, BCEA, BACE, BAEC, BECA, BEAC.
CAEB, CABE, CEAB, CEBA, CBAE, CBEA.
ABCE, ABEC, AEBC, AECB, ACBE, ACEB.
EBCA, EBAC, ECBA, ECAB, EABC, EACB.

Total 4 character words from 2 different consonants and 2 different vowels would be 9x24 = 216.

Apply to above.

7C3 x 4C2 x 5! = 25200.

More difficult would be when repetitions are allowed in consonants and vowels.

How many words of 4 consonants and 3 vowels can be?

How many words of 3 consonants and 3 vowels can be formed from 8 consonants and 4 vowels?

How many words can be formed by 2 vowels and 3 consonants out of 4 vowels and 7 consonants?

How many words of 3 consonants and 2 vowels can be formed from 5 consonants and 4 vowels?