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Important Formula and Tips Permutation & Combination GATE 2015

1.Factorial Notation:

Let n be a positive integer. Then, factorial n, denoted n! is defined as:

n! = n(n - 1)(n - 2) ... 3.2.1.

Examples:

We define 0! = 1.

4! = (4 x 3 x 2 x 1) = 24.

5! = (5 x 4 x 3 x 2 x 1) = 120.

Permutations:

The different arrangements of a given number of things by taking some or all at a time, are called permutations.

Examples:

All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).

All permutations made with the letters a, b, c taking all at a time are:
( abc, acb, bac, bca, cab, cba)

Number of Permutations:

Number of all permutations of n things, taken r at a time, is given by:

nPr = n(n - 1)(n - 2) ... (n - r + 1) = n!
(n - r)!
Examples:

6P2 = (6 x 5) = 30.

7P3 = (7 x 6 x 5) = 210.

Cor. number of all permutations of n things, taken all at a time = n!.

An Important Result:

If there are n subjects of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of third kind and so on and pr are alike of rth kind,
such that (p1 + p2 + ... pr) = n.

Then, number of permutations of these n objects is = n!
(p1!).(p2)!.....(pr!)
Combinations:

Each of the different groups or selections which can be formed by taking some or all of a number of objects is called a combination.

Examples:

Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA.

Note: AB and BA represent the same selection.

All the combinations formed by a, b, c taking ab, bc, ca.

The only combination that can be formed of three letters a, b, c taken all at a time is abc.

Various groups of 2 out of four persons A, B, C, D are:

AB, AC, AD, BC, BD, CD.

Note that ab ba are two different permutations but they represent the same combination.

Number of Combinations:

The number of all combinations of n things, taken r at a time is:

nCr = n! = n(n - 1)(n - 2) ... to r factors .
(r!)(n - r)! r!
Note:

nCn = 1 and nC0 = 1.

nCr = nC(n - r)

Examples:

i. 11C4 = (11 x 10 x 9 x 8) = 330.
(4 x 3 x 2 x 1)
ii. 16C13 = 16C(16 - 13) = 16C3 = 16 x 15 x 14 = 16 x 15 x 14 = 560.
3! 3 x 2 x 1

Concept of the Day GATE 2015 Preparation

Advantages of DC electric power transmission:
1. It requires only two conductors.

2. There is no problem of inductance, capacitance and phase displacement which is common in ac transmission.

3. For the same load and sending end voltage, the voltage drop in dc transmission lines is less than that in ac transmission.

4. As there is no skin effect on conductors, therefore entire cross-section of conductor is usefully utilized thereby affecting saving in material.

5. For the same value of voltage insulating materials on dc lines experience less stress as compared to those on ac transmission lines.

6. A dc line has less corona loss and reduced interference with communication circuits.

7. There is no problem of system instability so common in ac transmission.

Advantages of AC electric power Transmission:
1. Power can be generated at high voltages as there is no commutation problem.

2. Ac voltages can be conveniently stepped up or stepped down.

3. High voltage transmission of ac power reduces losses.

Topic - GATE 2015 (Digital Circuit) ECE/EE/CS

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Important Topic - GATE 2015 (Bessel Function) ECE

c(t) and m(t) are used to generate an FM signal. If the peak frequency deviation of the generated FM signal is three times the transmission bandwidth of the AM signal, then the coefficient of the term cos [2p(1008 x 103t)] in the FM signal (in terms of the Bessel Coefficients) is

A. 5 J4(3)
B. J8(3)
C. J8(4)
D. 5 J4(6)
As, we known FM, s(t)



Option D

Trigonometric identities and formulae

# Logic GATE

Question of the day GATE 2015 - Electronics/Computer science/Electrical engineering
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Minute Preperation Tips: Complete one subject in 3 Days - GATE 2015

Timer 555

24 Days to go- GATE 2015. Solve at-least 30 Numerical Today.

Formula of the day GATE 2015 - Electronics/Computer science/Electrical engineering

Formula of the day GATE 2015 - Mechanical engineering

Tips for GATE Exam 2015

1. Read carefully and understand the question completely. understand what is given and what is asked .

2. Try to make some headway with this information. for instance you dont remember or know a formula but know that the answer is directly proportional to something or inversely proportional to something else. try to formalize this partial knowledge in a helpful way

3. Look at the answer choices and eliminate ones that are obviously wrong
4. Look for patterns in the choices. sometimes, the choice that sticks out the most is probably a wrong choice. sometimes, all the choices contain the same variables but in different positions (numerator, denominator, exponent, log etc)

5. As a last resort and not as an excuse for not preparing but you will be surprised at how many questions you can answer if you follow this. This way of answering is dangerous and is an art. this art can be honed only attempting many practice questions of multiple choice types and not by just reading the theory. (Keyword is Guess)