How many committees of five people can be chosen from 20 men and 12 women such that each committee contains at least three women?
A. | 75240 |
B. | 52492 |
C. | 41800 |
D. | 9900 |
Option: B Explanation : We must choose at least 3 women, so we calculate the case of 3 women, 4 women and 5 women and by addition rule add the results. 12C3 x 20C2 + 12C4 x 20C1 + 12C5 x 20C0 = (12x11x10/3x2x1) x (20x19/2x1) + (12x11x10x9/4x3x2x1) x 20 + (12x11x10x9x8/5x4x3x2x1) x 1 = 220 x 190 + 495 x 20 + 792 = 52492 Click on Discuss to view users comments. |
Which of the following statement(s) is/are false?
(a) A connected multigraph has an Euler Circuit if and only if each of its vertices has even degree.
(b) A connected multigraph has an Euler Path but not an Euler Circuit if and only if it has exactly two vertices of odd degree.
(c) A complete graph (Kn) has a Hamilton Circuit whenever n≥3
(d) A cycle over six vertices (C6) is not a bipartite graph but a complete graph over 3 vertices is bipartite.
A. | (a) only |
B. | (b) and (c) |
C. | (c) only |
D. | (d) only |
Option: D Explanation : Click on Discuss to view users comments. |
The inorder traversal of the following tree is:
A. | 2 3 4 6 7 13 15 17 18 18 20 |
B. | 20 18 18 17 15 13 7 6 4 3 2 |
C. | 15 13 20 4 7 17 18 2 3 6 18 |
D. | 2 4 3 13 7 6 15 17 20 18 18 |
Option: D Explanation : Click on Discuss to view users comments. |
Which of the following is/ are not true ?
(a) The set of negative integers is countable.
(b) The set of integers that are multiples of 7 is countable.
(c) The set of even integers is countable.
(d) The set of real numbers between 0 and Ih is countable.
A. | (a) and (c) |
B. | (b) and (d) |
C. | (b) only |
D. | (d) only |
Option: D Explanation : Click on Discuss to view users comments. |
Consider the graph given below:
The two distinct sets of vertices, which make the graph bipartite are:
A. | (v 1 , v 4 , v 6 ); (v 2 , v 3 , v 5 , v 7 , v 8 ) |
B. | (v 1 , v 7 , v 8 ); (v 2 , v 3 , v 5 , v 6 ) |
C. | (v 1 , v 4 , v 6 , v 7 ); (v 2 , v 3 , v 5 , v 8 ) |
D. | (v 1 , v 4 , v 6 , v 7 , v 8 ); (v 2 , v 3 , v 5 ) |
Option: C Explanation :
A simple graph G=(V,E) is called bipartite if its vertex set can be partitioned into two disjoint subsets V=V 1 ⋃V 2 , such that every edge has the form e=(a,b) where aϵV 1 and bϵV 2 .
Bipartite graphs are equivalent to two-colorable graphs.
1. Assign Red color to the source vertex (putting into set V 1 ).
2. Color all the neighbours with Black color (putting into set V 2 ).
3.Color all neighbour’s neighbour with Red color (putting into set V 1 ).
4. This way, assign color to all vertices such that it satisfies all the constraints of m way coloring problem where m= 2.
5. While assigning colors, if we find a neighbour which is colored with same color as current vertex, then the graph cannot be colored with 2 colors (ie., graph is not Bipartite).
So answer is option (C).
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