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# 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!}}$

Last updated date: 20th Jun 2024
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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.

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.
$\Rightarrow 2 \times 2 \times 2 \times 2 \times ................ \times 10{\text{ times}}$
$\Rightarrow {2^{10}} = 1024$
Therefore, the option $\left( b \right)$ is correct.