This is precisely the problem we saw to solve the r-combination with repetition:
= C(8+4-1,8)
= ¹¹ C​ ₈
= 11! / (8!(11-8)!)
= 165

Waterfall model→ Automate the manual system for student record maintenance in a school
Incremental development→ e-business that starts with only the basic functionality and then moves on to more advanced features.
Prototyping→ A virtual reality system for simulating vehicle navigation in a highway.
RAD→ An inventory control system for a supermarket to be developed in a highway

Each process needs 2 drives. So for deadlock just give each process one drive. So total 6 process can be given 1 drive each and can cause deadlock. So to break deadlock just reduce 1 process.
So maximum no. of process for the system to be deadlock free is 5.

Total possible IP numbers numbers of all classes 0 to 255.
Class-A: 0 to 127. It means 50%
Class-B: 128 to 191. It means 25%
Class-C: 192 to 223. It means 12.5 %
Class-D: 224 to 239. It means 6.25%
Class-E: 240 to 255. It means 6.25%

Foreign key is a key whose values depends on the primary key of a relation. So for the functioning of a foreign key in the context of relational database we need a primary key.

Software Reuse is the process of existing software artifacts and knowledge to build new software.

Defuzzication Methods: Following defuzzication methods are known to calculate crisp output → Maxima Methods
1.Height method
2.First of maxima (FoM)
3.Last of maxima (LoM)
4.Mean of maxima(MoM)
→ Centroid methods:
1. Center of gravity method (CoG)
2. Center of sum method (CoS)
3. Center of area method (CoA)
→ Weighted average method

Take variable i=4*2.14
We are simply folding the value into 8.56 because to avoid multiplication costly operation.

Symbol tables are data structures that are used by compilers to hold information about source-program constructs. The information is collected incrementally by the analysis phases of a compiler and used by the synthesis phases to generate the target code. Entries in the symbol table contains information about an identifier such as its character string (or lexeme) , its type,its position in storage, and any other relevant information.

Excluded for evaluation

There are 2​ ⁿ​ different n-tuples of 0s and 1s.
A Boolean function is an assignment of 0 or 1 to each of these 2​ ⁿ​ different n-tuples.
Therefore, there are 2​ ^{2^n} different Boolean functions.
Example: How many different Boolean functions of degree 4 are there?
Solution: 16

(146)₇+(317)_7-2=(246)₈
Substitute 7 in b,
(146)₇+(317)_7-2=(246)₈
(146)₇+(317)₅=(246)₈
(146)₇=1*7²+4*7¹+6*7⁰
=49+28+6
=83
(317)₅=3*5²+1*5¹+7*5⁰
=75+5+7
= 87
(246)₈=2*8²+4*8¹+6*8⁰
=128+32+6
= 166
∴ (146)₇+(317)₅=(246)₈
=83+87
=166
166=166
LHS = RHS equal only if b is 7.

Disk--> File system
CPU → Thread
memory → Virtual address space
Interrupt → Signal

encryption and decryption are functions of presentation layer in the OSI reference model. Here, they were given TCP/IP model. It combines Session, presentation and Application layers into Application layer.

We apply Euler’s formula where r = e−v + 2.
Since each vertex has degree 4, the sum of the degrees is 24.
By the handshaking theorem, 2e = 24 .
so, e = 12.
r = 12−6 + 2
r = 8
Thus we have 8 regions in this planar graph.

(a): L(G)=∅ is a emptiness problem and emptiness problem for context free languages is decidable.

(b): Recursive: The reason is we can assume L(M2)=∑* and hence any TM (M1) will be definitely subset of L(M2). So we only need to see that M1 is a valid Turing machine and no need to check whether it is subset of L(M2) as it is obviously subset of L(M2)=∑*. Checking validity of Turing machine is always a halting process, hence it is recursive.

(c): L={(M,w)|M is a TM that accepts w using at most 2^|w| squares of its tape}.
Recursive: Let m be he number of states in M, and k be the size of the alphabet that M uses, and r=|w|. If M uses at most 2^r squares of its tape, then there are at most Alpha=mK^2r 2^r configurations(why?). If M runs on w for more than α steps, and does not use more than 2^r squares of its tape, then M must be in the one configuration at least twice(pigeonhole principle), in which case M would enter an infinite loop on input w. Based on this, we design a machine M* that decides L that works as follows:
M* runs M on w for at most α+1 steps.
If M accepts w which steps with using at most 2^r squares, M* halts and accepts.
If M rejects w within α steps with using at most 2^r squares, M* halts and rejects.
If M goes beyond 2^r squares, M* halts and rejects.
If M does not halt and does not go beyond 2^r squares, M* rejects.
Note: To know the definition of configurations and the details, reader may refer to "Introduction to theory of computation Michael Sipser, Chapter 8 Space complexity)

Φ(45) → Φ(9*5)
→ Φ(3² * 5)
→ Φ(3²) * Φ(5)
→ Φ(3² - 3,^(2-1)) * (5-1)
= 24
Remember Formula:
If p is prime then Φ(p) = (p-1)
and if p is not prime and it's in prime power k from then Φ(p^k) = p^k - p(k-1)

Given h(k,i)=h1(k)+ih2(k)
where h1(k)=k mod m
h2(k)=1+k mod n
where n=m-1 and m=701. For k=123456
h1(k)=123456 mod 701
h1(k)=80
h2(k)=1+(123456 mod 700)
h2(k)=1+256=257
1st probe: i=1
h(k,i)=h1(k)+ih2(k)
h(k,1)=h1(k)+ih2(k)
h(k,1)=80+257=337
h(k,1)=337
2nd probe: i=2
h(k,2)=80+2257
h(k,2)=80+514
h(k,2)=594
∴ Difference h(k,2)-h(k,1)
594-337
257