Among 9 consonants and 5 vowels, what number of words can be formed of 4 consonants and 3 vowels?

From $5$ consonants and $4$ vowels, how many words can be formed by using $3$ consonants and $2$ vowels.A. 9440 B. 6800 C. 3600D. 7200

Answer

Hint: The number of ways a word can form from $5$ consonants by using $3$ consonants $ = $ ${}^5{C_3}$ and from $4$ vowels by using $2$ vowels $ = $${}^4{C_2}$, hence the number of words can be $ = {}^5{C_3} \times {}^4{C_2} \times {}^5{P_5}$. Use this to find the no. of words.

Complete step-by-step solution:
According to the question it is given that :
From$5$consonants , $3$ consonants can be selected and from $4$ vowels , $2$ vowels can be selected .
So, from $5$ consonants , $3$ consonants can be selected in ${}^5{C_3}$ ways.
From $4$ vowels ,$2$ vowels can be selected in ${}^4{C_2}$ways.
Now with every selection , the number of ways of arranging $5$ letters in ${}^5{P_5}$ways.
Hence, total number of words $ = {}^5{C_3} \times {}^4{C_2} \times {}^5{P_5}$
$\therefore $we know that
$
  {}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} \\
  {}^n{P_r} = \dfrac{{n!}}{{(n - r)!}} \\
$
Hence , total number of words $ = {}^5{C_3} \times {}^4{C_2} \times {}^5{P_5}$
                                                       $
   = \dfrac{{5!}}{{3!(5 - 3)!}} \times \dfrac{{4!}}{{2!(4 - 2)!}} \times \dfrac{{5!}}{{(5 - 5)!}} \\
   = \dfrac{{5 \times 4 \times 3!}}{{3! \times 2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2! \times 2!}} \times 5! \\
   = 5 \times 2 \times 2 \times 3 \times 120 \\
   = 7200 \\
$

Note: It is advisable in such types of questions we should see that what are all possibilities that words can be formed , for this one must have a basic understanding of permutation and combination. Here we have used $ {}5{P_5}$ for arranging 5 words.

From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if any letter can be repeated?

Your answer $7^4 \cdot 5^3$ is the number of ways the first four positions can be filled with consonants and the last three positions can be filled with vowels when letters may be repeated. However, the problem does not specify the positions of the consonants and vowels, so we must select them.

Notice that selecting which four of the seven positions will be filled with consonants also determines the positions of the three vowels since each of the remaining three positions must be filled with vowels.

We must select which four of the seven positions must be filled with consonants, which can be done in $\binom{7}{4}$ ways. Each of those four selected positions can be filled with a consonant in seven ways. Each of the remaining three positions can be filled with a vowel in five ways. Hence, there are $$\binom{7}{4}7^4 \cdot 5^3$$ words that can be formed consisting of four consonants and three vowels selected from an alphabet of seven consonants and three vowels with repetition.

The stated answer $7^7\dbinom{7}{4}\dbinom{5}{3}$ is nonsense.

From $7$ consonants and $5$ vowels, how many words can be formed consisting of $4$ consonants and $3$ vowels if no letter can be repeated?

Your answer $P(7, 4)P(5, 3)$ is the number of ways of filling the first four positions with distinct consonants and the last three positions with distinct vowels. However, the problem does not specify the positions of the consonants and vowels, so we must select them.

As above, we select four of the seven positions for the consonants. There are $P(7, 4)$ ways to arrange four of the seven consonants in those positions and $P(5, 3)$ ways to arrange three of the five vowels in the remaining three positions. Hence, there are $$\binom{7}{4}P(7, 4)P(5, 3)$$ words that can be formed consisting of four distinct consonants and three distinct vowels selected from an alphabet of seven consonants and five vowels.

Notice that $$\binom{7}{4}P(7, 4)P(5, 3) = \frac{7!}{4!3!} \cdot \frac{7!}{3!} \cdot \frac{5!}{2!} = 7! \cdot \frac{7!}{4!3!} \cdot \frac{5!}{2!3!} = 7!\binom{7}{4}\binom{5}{3}$$ The author of your book arrived at the answer by selecting four of the seven consonants in $\binom{7}{4}$ ways, selecting three of the five vowels in $\binom{5}{3}$ ways, and then arranging the seven distinct selected letters in $7!$ ways.

How can 7 consonants and 4 vowels words of three consonants and two vowels can be formed?

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

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

How many words of 3 consonants and 3 vowels can be formed from 6 consonants and 5 vowels?