Monday 11 April 2016

CBSE-UGC NET RESULT - December 2015

http://cbseresults.nic.in/UGC/net_dec2015.htm

NATIONAL ELIGIBILITY TEST (NET)

On behalf of UGC, the Central Board of Secondary Education announces holding of the National Eligibility Test (NET) on 10th July, 2016 (SUNDAY) for determining the eligibility of Indian nationals for the Eligibility for Assistant Professor only or Junior Research Fellowship & Eligibility for Assistant Professor Both in Indian universities and colleges. CBSE will conduct NET in 83 subjects at 88 selected NET Examination Cities spread across the country.
The candidates who qualify for the award of Junior Research Fellowship are eligible to pursue research in the subject of their post-graduation or in a related subject and are also eligible for Assistant Professor. The universities, institutions, IITs and other national organizations may select the JRF awardees for whole time research work in accordance with the procedure prescribed by them. The award of JRF and Eligibility for Assistant Professor both OR Eligibility for Assistant Professor only will depend on the performance of the candidate in all three papers of NET. However, the candidates qualifying exclusively for Assistant Professor will not be considered for award of JRF.

Tuesday 5 April 2016

UGC NET JULY 2016 Notification





E-mail: net@cbse.gov.in

Website: www.cbsenet.nic.in

Tele. 7042399520, 7042399521,

7042399524, 7042399525,

7042399526, 7042399528,

7042399529.

Fax. 0120-2427772

CENTRAL BOARD OF SECONDARY EDUCATION

(An Autonomous Organisation under the Union Ministry of Human Resource Development Govt. of India)

H-149, Sector-63, Noida, District Gautam Budh Nagar – 201 309(UP).

_________________________________________________________________________

CBSE/NET/2015-16 04/04/2016

UGC NATIONAL ELIGIBILITY TEST (NET)

SHORT NOTIFICATION

It is notified that CBSE will conduct the next UGC-NET for Junior Research

Fellowship & Eligibility for Assistant Professor on 10th July 2016 (Sunday). The candidates who

desire to appear in the test may see the detailed notification available on the website

www.cbsenet.nic.in from 12th April, 2016. The candidates are required to apply online from

12.04.2016. The last date for applying online is 12st May 2016 and fee can be paid up to 13

May, 2016.

th

EXECUTIVE DIRECTOR

Download Complete notification here


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Saturday 2 April 2016

UGC-NET Computer Science Data Structures Questions with Explanation

In this Blog we are providing all the UGC-NET Computer Science previous year Questions with explanation:

Q::7 Number of binary trees formed with 5
nodes are
(A) 32 (B) 36
(C) 120 (D) 42


Answer:(D)
Explanation:
In combinatorial mathematics, the Catalan numbers form a sequence of natural numbers that occur in various counting problems, often involving recursively-defined objects. They are named after the Belgian mathematician Eugène Charles Catalan (1814–1894).
Using zero-based numbering, the nth Catalan number is given directly in terms of binomial coefficients by
C_n = \frac{1}{n+1}{2n\choose n} = \frac{(2n)!}{(n+1)!\,n!} = \prod\limits_{k=2}^{n}\frac{n+k}{k} \qquad\mbox{ for }n\ge 0.
The first Catalan numbers for n = 0, 1, 2, 3, … are
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, … 


Applications in combinatorics

There are many counting problems in combinatorics whose solution is given by the Catalan numbers. The book Enumerative Combinatorics: Volume 2 by combinatorialist Richard P. Stanley contains a set of exercises which describe 66 different interpretations of the Catalan numbers. Following are some examples, with illustrations of the cases C3 = 5 and C4 = 14.

Lattice of the 14 Dyck words of length 8 - ( and ) interpreted as up anddown
  • Cn is the number of Dyck words of length 2n. A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's (see also Dyck language). For example, the following are the Dyck words of length 6:
XXXYYY     XYXXYY     XYXYXY     XXYYXY     XXYXYY.
  • Re-interpreting the symbol X as an open parenthesis and Y as a close parenthesis, Cn counts the number of expressions containing n-pairs of parentheses which are correctly matched:
((()))     ()(())     ()()()     (())()     (()())
  • Cn is the number of different ways n + 1 factors can be completely parenthesized (or the number of ways of associating n applications of a binary operator). For n = 3, for example, we have the following five different parenthesizations of four factors:
((ab)c)d     (a(bc))d     (ab)(cd)     a((bc)d)     a(b(cd))

The associahedron of order 4 with the C4=14 full binary trees with 5 leaves
  • Successive applications of a binary operator can be represented in terms of a full binary tree. (A rooted binary tree is full if every vertex has either two children or no children.) It follows that Cn is the number of full binary trees with n + 1 leaves:
Catalan number binary tree example.png
  • Cn is the number of non-isomorphic ordered trees with n vertices. (An ordered tree is a rooted tree in which the children of each vertex are given a fixed left-to-right order.)
  • Cn is the number of monotonic lattice paths along the edges of a grid with n × n square cells, which do not pass above the diagonal. A monotonic path is one which starts in the lower left corner, finishes in the upper right corner, and consists entirely of edges pointing rightwards or upwards. Counting such paths is equivalent to counting Dyck words: X stands for "move right" and Y stands for "move up".
The following diagrams show the case n = 4:
Catalan number 4x4 grid example.svg
This can be succinctly represented by listing the Catalan elements by column height:
[0,0,0,0][0,0,0,1][0,0,0,2][0,0,1,1]
[0,1,1,1] [0,0,1,2] [0,0,0,3] [0,1,1,2][0,0,2,2][0,0,1,3]
[0,0,2,3][0,1,1,3] [0,1,2,2][0,1,2,3]

The triangles correspond to nodes of the binary trees.
  • Cn is the number of different ways a convex polygon with n + 2 sides can be cut into triangles by connecting vertices with straight lines(a form of Polygon triangulation). The following hexagons illustrate the case n = 4:
Catalan-Hexagons-example.svg
  • Cn is the number of stack-sortable permutations of {1, ..., n}. A permutation w is called stack-sortable if S(w) = (1, ..., n), where S(w) is defined recursively as follows: write w = unv where n is the largest element in w and u and v are shorter sequences, and set S(w) = S(u)S(v)n, with S being the identity for one-element sequences. These are the permutations that avoid the pattern 231.
  • Cn is the number of permutations of {1, ..., n} that avoid the pattern 123 (or any of the other patterns of length 3); that is, the number of permutations with no three-term increasing sub-sequence. For n = 3, these permutations are 132, 213, 231, 312 and 321. For n = 4, they are 1432, 2143, 2413, 2431, 3142, 3214, 3241, 3412, 3421, 4132, 4213, 4231, 4312 and 4321.
  • Cn is the number of non-crossing partitions of the set {1, ..., n}. A fortioriCn never exceeds the nth Bell number. Cn is also the number of non-crossing partitions of the set {1, ..., 2n} in which every block is of size 2. The conjunction of these two facts may be used in a proof by mathematical induction that all of the free cumulants of degree more than 2 of the Wigner semicircle law are zero. This law is important in free probability theory and the theory of random matrices.
  • Cn is the number of ways to tile a stairstep shape of height n with n rectangles. The following figure illustrates the case n = 4:
Catalan stairsteps 4.svg
  • Cn is the number of rooted binary trees with n internal nodes (n + 1 leaves or external nodes). Illustrated in following Figure are the trees corresponding to n = 0,1,2 and 3. There are 1, 1, 2, and 5 respectively. Here, we consider as binary trees those in which each node has zero or two children, and the internal nodes are those that have children.
Binary Tree.png