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With $4$ bits there are $2^4=16$ combinations. How many combinations can you have with $4$ qubits (assuming they are all superimposed)?

**EDIT:**

As per here, the argument is that if a qubit can hold more information when it's measurement(probability) is collapsed to give a $|1\rangle$ or a $|0\rangle$ (assuming measurement is without errors). For a superimposed qubit holding two values at the same time depending on how it has been encoded i.e. as per the parallel measurement (probably in a quantum gate) we could get $|00\rangle$, $|01\rangle $, $|10\rangle$, $11\rangle$. I am seeing this as if observed in parallelism 4 qubits would yield a combination ($2^8$) of 8 classical bits (i.e. n qubits having $2^{n+n}$ number of combinations in parallel). Is this correct?

1Hi @LiNKer! Welcome to QCSE! This question is a little broad right now, because it's unclear what's being asked and what you know of. Nonetheless, $4$ qubits can also be in a uniform superposition of all $2^{4}=16$ basis states; a difference between a "normal pc" (what is called a classical computer) and a quantum computer is that the bits in a classical computer are

either$0$ or $1$, while for a quantum computer the qubits can be in asuperpositionof $0$ and $1$. Can you consider revising your question to provide more clarity on what you know of and where your question comes from? – Mark S – 2019-09-25T17:18:45.980Hi Mark, thanks for the suggestion. made the edit – LiNKeR – 2019-09-25T17:30:51.307

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Possible duplicate of Understanding (theoretical) computing power of quantum computers

– ahelwer – 2019-09-25T20:37:07.293