The intrinsic charge carrier density in germanium crystal at 300 K is $ 2.5 \times $ $ 10^{13} / \mathrm{cm}^{3} . $ density in an n- type germanium crystal at $ 300 \mathrm{K} $ be $ 5 \times 10^{16} / \mathrm{cm}^{3} $ ,the hole density in this n-type crystal at $ 300 \mathrm{K} $ would be
(A) $ \quad 2.5 \times 10^{13} / \mathrm{cm}^{3} $
(B) $ \quad 5 \times 10^{6} / \mathrm{cm}^{3} $
(C) $ \quad 1.25 \times 10^{10} / \mathrm{cm}^{3} $
(D) $ \quad 0.2 \times 10^{4} / \mathrm{cm}^{3} $

Hint
We know that semiconductors are materials which have a conductivity between conductors which are generally metals and non-conductors or insulators such as most ceramics. Semiconductors can be pure elements, such as silicon or germanium, or compounds such as gallium arsenide or cadmium selenide. Semiconductors are employed in the manufacture of various kinds of electronic devices, including diodes, transistors, and integrated circuits. Such devices have found wide application because of their compactness, reliability, power efficiency, and low cost. On the basis of this we have to solve this question.

Complete step by step answer
Let us first explain the kinds of semiconductors.
So, an intrinsic semiconductor is an undoped semiconductor. This means that holes in the valence band are vacancies created by electrons that have been thermally excited to the conduction band, as opposed to doped semiconductors where holes or electrons are supplied by a foreign atom acting as an impurity. On the other hand, an extrinsic semiconductor is a semiconductor doped by a specific impurity which is able to deeply modify its electrical properties, making it suitable for electronic applications on diodes, transistors, and many other or optoelectronic applications like on light emitters and detectors.
So, we can write that
From the law of mass action, law states that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the equation.
 $ nP=n{{i}^{2}} $
So, after we put the values, we get:
 $ \mathrm{n}=\dfrac{2.5 \times 2.5 \times 10^{26}}{5 \times 10^{16}} $
 $ \mathrm{n}=1.25 \times 10^{10} / \mathrm{cm}^{3} $.
Hence, the option (C) is the correct answer.

Note
We know that the positive charge carriers such as holes are the charge carriers that carry positive charge with them while moving from one place to another place. Holes are the vacancies in the valence band that move from one place to another place within the valence band. Free carriers are electrons or holes which have been introduced directly into the conduction band or valence band by doping and are not promoted thermally. For this reason, electrons holes will not act as double carriers by leaving behind holes electrons in the other band.
Thus, we can say that a charge carrier is a particle or quasiparticle that is free to move, carrying an electric charge, especially the particles that carry electric charges in electrical conductors.