madeeasy ots digital logic

Question

In the 2^nd cycle as it is ripple counter Q0 will change and it will 0 so for that reason Q1 will not get clock and it will be in previous state that is 1. So output should be 01. Right??

Comments

Yes, you are correct. After 2^nd Clock $Q_{0} = 0$  and   $Q_{1} = 1$  so  $Q_{0}Q_{1} = 01 $

But here the sequence is asking $Q_{1}Q_{0}$ , not  $Q_{0}Q_{1}$ so its $10$  and the solution given according to that. 

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Yea I understand silly mistakes.