Question

Find the number of n digit numbers formed using the first 5 natural numbers, which contain the digits 2 & 4 essentially

Answer 1

Let us consider multiple cases to solve this:

Total number of numbers present of $n$ digit made up of $1^{st}$ five natural numbers is $S=5^n$

case$1$ This case considers with atleast one $2$ present and no $4$'s present.

$\Rightarrow S_0={{}^{n}C}_1{3}^{n-1}+{{}^{n}C}_2{3}^{n-2}+{{}^{n}C}_3{3}^{n-3}+....+{{}^{n}C}_n$

$\Rightarrow S_0=4^n-1$

Case$2$ This considers with atleast one $4$ present and no $2$'s present.

$\Rightarrow S_1={{}^{n}C}_1{3}^{n-1}+{{}^{n}C}_2{3}^{n-2}+{{}^{n}C}_3{3}^{n-3}+....+{{}^{n}C}_n$

$\Rightarrow S_1=4^n-1$

Case$3$ No $2$'s or $4$'s are present.

$\Rightarrow S_2=3^n$

So, total number of such values are

$s=S-S_0-S_1-S_2=5^n-2(4^n-1)-3^n$