Question

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

Answer 1

First $5$ natural numbers $=1,2,3,4,5$

Now we have to take $2$ and $4$ essentially that means minimum $n$ is $2$ and maximum will be $5.$

$\text{Considering digits to be non repetitive.}$

So, number of $2$ digit numbers is $2$ since only $24$ and $42$ are possible as it is must to take these digits

$3$ digit numbers$:-$

By permutation and combination:-

When $1$ is chosen as 3rd digit:-

$\Rightarrow 3!$ numbers are possible

Similarly with $3,5$

So total $3$ digit numbers possible $=6+6+6$ $=18$

$4$ digit numbers:-

Combinations of choosing among $1,3,5{=}^3{\text{C}}_2$

Now, permutations possible for all $3$ combinations $=4!\times 3$

Total $4$ digit numbers $=72$

$5$ digit numbers:-

Combination possible among $1,3,5={}^3{\text{C}}_3$

Now, permutation possible for all combination $=5!\times 1$

Total $5$ digit numbers $=120$

So total numbers possible $=2+18+72+120=212$