Register

UGC NET CSE | September 2013 | Part 2 | Question: 18

• recategorized by
5,322 views

1 Answer

Best answer
8 votes
8 votes
Number of 8-bit strings beginning with 111 are 32. First 3 bits are fixed and the remaining 5 bits can be 0 or 1. So the total combinations are $2^5=32$. Same is the case with the strings starting with 101. So total number of strings are $32+32=64$.
• selected by
User Avatar
Voting & Accepting an answer helps improve the community
Answer:
← PreviousNext →
← Previous in categoryNext in category →

Related questions

1 votes
1 votes
1 answer
2
go_editor asked Jul 20, 2016
1,545 views
UGC NET CSE | September 2013 | Part 2 | Question: 16
LL grammar for the language $L = \{a^n b^m c^{n+m} \mid m \geq 0, n \geq 0\}$ is $ S \rightarrow aSc \mid S_1 ; S_1 \rightarrow bS_1c \mid \lambda$ ... $ S \rightarrow aSc \mid S_1 \lambda ; S_1 \rightarrow bS_1c \mid \lambda$
LL grammar for the language $L = \{a^n b^m c^{n+m} \mid m \geq 0, n \geq 0\}$ is$ S \rightarrow aSc \mid S_1 ; S_1 \rightarrow bS_1c \mid \lambda$$ S \rightarrow aSc \mid...
User Avatar
4 votes
4 votes
3 answers
4
go_editor asked Jul 20, 2016
1,269 views
UGC NET CSE | September 2013 | Part 2 | Question: 14
Regular expression for the language $L=w \in \{0,1\}* \mid w$ has no pair of consecutive zeros $\}$ is $(1+010)*$ $(01+10)*$ $(1+010)*(0+ \lambda)$ $(1+01)* (0+\lambda)$
Regular expression for the language $L=w \in \{0,1\}* \mid w$ has no pair of consecutive zeros $\}$ is$(1+010)*$$(01+10)*$$(1+010)*(0+ \lambda)$$(1+01)* (0+\lambda)$
User Avatar
Link Copied!