Question

Let $C$ be a set of $6$ consonants $\{b,c,d,f,g,h\}$ and $V$ be the set of $5$ vowels $\{a,e,i,o,u\}$ and $W$ be the set of seven-letter words that can be formed with these $11$ letters using both the following rules.
$\text{I}.$ The vowels and consonants in the word must alternate.
$\text{II}.$ No letter can be used more than once in a single word.

Then the number of words in the set $W$ is

• A. $32000$
• B. $36000$
• C. $42000$
• D. $38000$

Answer 1

The correct option is B $36000$

Consonants $\to \{b,c,d,f,g,h\}$
Vowels $\to \{a,e,i,o,u\}$
$$\begin{array}{ccccccc}\times & & \times & & \times & & \times \end{array}$$
Case $1:$ If the word begins with a consonant, then
Number of words $={(}^6{C}_4\times 4!)\times {(}^5{C}_3\times 3!)$
$=360\times 60=21600$

Case $2:$ If the word begins with a vowel, then
Number of words $={(}^5{C}_4\times 4!)\times {(}^6{C}_3\times 3!)$
$=120\times 120=14400$

$\therefore$ The number of words in set $W$ equals $21600+14400=36000$