Question

The number of seven-digit natural numbers in which only $2$ and $3$ are present as digits and no two $2^{\prime }s$ are consecutive in any number, is

• A. $26$
• B. $34$
• C. $32$
• D. $53$

Answer 1

The correct option is B $34$

No two $2^{\prime }s$ should be consecutive.
Case $1:$ No $2^{\prime }s$ occur.
Count of such number $=1$

Case $2:$ Only one $2^{\prime }s$ occurs.
Count of such number $={}^7{C}_1=7$

Case $3:$ Two $2^{\prime }s$ occur.
$\underset{–}{x_1}\underset{–}{2}\underset{–}{x_2}\underset{–}{2}\underset{–}{x_3}$
Arrangement can be done in this way, where $x_1\ge 0,x_2\ge 1,x_3\ge 0$
and $x_1+x_2+x_3=5$
Putting $x_2^{\prime }=x_2-1$ so that $x_2^{\prime }\ge 0$
$x_1+x_2^{\prime }+x_3=4$
Number of solutions $={}^{4+3-1}{C}_{3-1}={}^6{C}_2=15$

Case $4:$ Three $2^{\prime }s$ occur.
$\underset{–}{x_1}\underset{–}{2}\underset{–}{x_2}\underset{–}{2}\underset{–}{x_3}\underset{–}{2}\underset{–}{x_4}$
$x_1+x_2+x_3+x_4=4$
$\Rightarrow x_1+x_2^{\prime }+x_3^{\prime }+x_4=2$
Number of solutions $={}^{2+4-1}{C}_{4-1}={}^5{C}_3=10$

Case $4:$ Five $2^{\prime }s$ occur.
Count of such number $=1$

Hence, total count $=1+7+15+10+1=34$