The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to _______. [2024]
(3734)
The word 'DISTRIBUTION' has total 12 letters and 9 diferent letters.
We select four different letters.
Case l : Three same letters and 1 different letter
Case 2 :Two pairs are same and two pairs are different
Case 3 : Two pair of letters are same
Case 4 : All different letters are select
Total letters selected.
There are 5 points on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points on the side BC and 7 points on the side CA of the triangle. The number of triangles, that can be formed using the points as vertices, is: [2024]
796
751
771
776
(2)
Number of points on side AB = 5
Number of points on side BC = 6
Number of points on side AC = 7
Number of ways of selecting three points from side AB =
Number of ways of selecting three points from side BC =
Number of ways of selecting three points from side AC =
Total number of triangles that can be formed using the points
The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is: [2024]
16
48
24
56
(1)
Total number of triangles that can be formed by using 8 vertices of the octagon =
Number of triangles having exactly one side common with the octagon =
Since, if we choose AB as the common side, then other vertices of the triangle will be either of G, F, E or D as we have exactly one common side.
So, we have i.e., such triangles.
Now, let us find the number of triangles having two common sides.
In the octagon, we have 8 ways of choosing two consecutive sides, i.e., (AB, BC), (BC, CD), (CD, DE), (DE, EF), (EF, FG), (FG, GH), (GH, HA), (HA, AB)
Number of triangles having 2 sides common with the octagon = 8
Required number of triangles =
If for some and then is equal to [2024]
372
384
380
376
(1)
Given,
i.e.,
or i.e., i.e.,
The number of ways, in which the letters A, B, C, D, E can be placed in the 8 boxes of the figure below so that no row remains empty and at most one letter can be placed in a box is :
[2025]
5880
5760
840
960
(2)
We have 5 letters A, B, C, D, E.
The following table shows the numbers of ways to fill 5 boxes so that no row remains empty.
Row-1 | Row-II | Row-III | Number of ways |
3 | 1 | 1 | |
2 | 2 | 1 | |
1 | 3 | 1 | |
2 | 1 | 2 | |
1 | 2 | 2 |
Number of ways to fill boxes = 6 + 18 + 6 + 9 + 9 = 48
Now, these 5 letters can be arranged in 5! ways
Total number of ways .
Line of slope 2 and line of slop intersect at the origin O. In the first quadrant, are 12 points on line and are 9 points on line . Then the total number of triangles, that can be formed having vertices at three of the 22 points O, is : [2025]
1080
1026
1134
1188
(3)
Total number of triangles are
= 594 + 432 + 108 = 1134
From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include atleast 4 batsmen and atleast 4 bowlers. One batsmen and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is [2025]
145
155
165
135
(2)
To select 10 players including atleast 4 batsmen and 4 bowlers.
Captain and vice-captain already selected
Total number of ways =
=
= 60 + 75 + 20 = 155
There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is [2025]
200
210
230
220
(2)
The number of ways to choose 3 points from 12 =
= = 220
Number of ways to choose 3 points from 5 collinear points = = 10
Since these 5 points are collinear, so they can not form any triangles.
Number of triangles that can be formed = 220 –10 = 210
For all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is 'M', is : [2025]
14950
5148
4356
6084
(2)
Number of English alphabets before M = 12
Number of English alphabets after M = 13
Required number of ways
= = 5148
Group A consists of 7 boys and 3 girls, while group B consists of 6 boys and 5 girls. The number of ways, 4 boys and 4 girls can be invited for a picnic if 5 of them must be from group A and the remaining 3 from group B, is equal to : [2025]
9100
8925
8750
8575
(2)
Group A : 7B and 3G
Group B : 6B and 5G.
The number of ways of inviting 4 boys and 4 girls (5 from Group A and 3 fro Group B) is gven by
= (3G + 2B, 1G + 2B) + (2G + 3B, 2G + 1B) + (1G + 4B, 3G + 0B)
= 1575 + 6300 + 1050 = 8925.