# In Qiskit, when creating multi-bit quantum and classical registers, what is the ordering of MSB to LSB?

When creating multi-bit quantum and classical registers, what is the ordering of MSB to LSB? For instance, I created a quantum register via

import qiskit
qr = qiskit.QuantumRegister(2, name='qr')


Is qr[1] the MSB, or is it qr[0]?

import qiskit
qr = qiskit.QuantumRegister(2, name='qr')
cr = qiskit.ClassicalRegister(2, name='cr')
qc = qiskit.QuantumCircuit(qr, cr)

# Result should be Bell state (|00>+|11>)/sqrt(2)
qc.h(qreg[0])
qc.cx(qreg[0], qreg[1])

# Result should be state |00>
qc.cx(qreg[0], qreg[1])
qc.h(qreg[0])

# Result should be state |10>
qc.x(qreg[0])

qc.measure(qreg, creg)
qc.draw()


At the end here I expect the quantum register to be in the state |10>, but the histogram below shows the state |01>. This indicates to me the MSB is the right-most bit, but I don't understand why the authors would do that. Is there some fundamental misunderstanding in my approach?

EDIT: Also the example is a bit convoluted--I initialized a Bell state because that's what the input to my circuit will be, but the same result happens with just the NOT gate on qr[0].

$$|00\rangle \rightarrow |01\rangle$$, which is "1" as an integer. $$1 = 2^0$$
$$|00000000\rangle \rightarrow |10000000\rangle = 32 = 2^7$$