Question & Answer
QUESTION

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

ANSWER Verified Verified
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.