Answer :

So, if we consider p-n-p transistor, the emitter will have holes as the majority charge carriers and they will move to the base due to forward biasing of collector-base junction. Very few holes will combine with the electrons of the base (since it is n-type) due to the thin size and light doping of the base. Because of reverse biasing at collector-base junction, majority of the holes will move to the collector side and constitute Ic which will be slightly less than Ie since no. of holes reaching the collector is less than that at emitter. Also, at the base some holes from the emitter can escape from the base terminal but due very thin size of the base a very little holes can only pass through terminal and constitute Ib and thus Ib is much smaller than Ie.

Thus, from explanation it can be seen that option A and C are correct.

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