Question

How many $10$ digits can be written by using digits $\left( {9{\text{ and 2}}} \right)$?
$\left( a \right){\text{ }}{}^{10}{{\text{c}}_1} + {}^9{{\text{c}}_2}$
$\left( b \right){\text{ }}{{\text{2}}^{10}}$
$\left( c \right){\text{ }}{}^{10}{{\text{c}}_2}$
$\left( d \right){\text{ 10!}}$

Answer 1

Hint: So for solving this question we must know permutation and combination. Here the case for one digit made of a nine or two which yields two possible answers. And with this, we can easily answer this question.

Complete step-by-step answer:
In one's place, there are two digits we can choose from$9{\text{ and }}2$. The same logic can be applied to the other 9 digits places because both $9{\text{ and }}2$ can be used as many times as necessary.
So the total number of numbers can be written by-
$ \Rightarrow 2 \times 2 \times 2 \times 2 \times ................ \times 10{\text{ times}}$
And it will be equal to
$ \Rightarrow {2^{10}} = 1024$
Therefore, the option $\left( b \right)$ is correct.

Additional information:
 Permutation means all possible arrangements of Number, letter, or any other product or things, etc. Well, the most basic difference in that permutations is ordered sets. That is the order of the elements matters for permutations. In combinations, the order is irrelevant, only the identity of the elements matters. The permutation is a course of action of things where the request for a game plan matters. The situation of everything in change matters. Consequently, Permutation can be related to Position.
The combination is gathering or determination of things where the request doesn't make a difference. Permutation can be considered as an arranged combination.

Note: In Permutation, arrangement matters in Combination, the arrangement does not matter. So while solving this type of problem we should have to keep in mind whether it is of arrangement or it is of grouping. So we should always be kept in mind while solving such types of problems so that we cannot get the error and can solve it easily without any problem.