Topic Question Set


Q 11 :    

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 C11×C18×4!3!

Case 2 :Two pairs are same and two pairs are different C12×C28×4!2!

Case 3 : Two pair of letters are same C22·C24

Case 4 : All different letters are select C49×4!

Total letters selected.

=C11×C18×4!3!+C12×C28·4!2!+C22·C24+C49·4!

=1×8!7!×4!3!+2!×8!6!2!×4!2!+1×4!2!2!+9!×4!5!4!

=32+672+6+3024=3734



Q 12 :    

There are 5 points P1,P2,P3,P4,P5 on the side AB, excluding A and B, of a triangle ABC. Similarly, there are 6 points P6,P7,,P11 on the side BC and 7 points P12,P13,,P18 on the side CA of the triangle. The number of triangles, that can be formed using the points P1,P2,,P18 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 = C35 

Number of ways of selecting three points from side BC = C36 

Number of ways of selecting three points from side AC = C37

Total number of triangles that can be formed using the points P1,P2,,P18=C318-(C3+C3+C3765)

=C3-C35-C36-C3718=816-10-20-35=751



Q 13 :    

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 = C38

Number of triangles having exactly one side common with the octagon = 8×4

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 8×4 i.e., 32 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 = C38-32-8

        =56328=16

 



Q 14 :    

If for some m,n; Cm6+2(Cm+16)+6Cm+2>8C3 and P3n-1:nP4=1:8, then Pm+1 n+ n+1Cm is equal to                    [2024]

  • 372

     

  • 384

     

  • 380

     

  • 376

     

(1)

Given, Cm6+2(Cm+16)+6Cm+2>8C3

6Cm+6Cm+1+6Cm+1+6Cm+2>8C3

7Cm+1+7Cm+2>8C38Cm+2>8C3 or 8C6-m>8C3

m+2>3 i.e., m>1

or 6-m>3 i.e., m<3 i.e., 1<m<3m=2

   Pm+1 n+ n+1Cm=8P3+9C2=336+36=372

 



Q 15 :    

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 C33·C13·C12=6
2 2 1 C23·C23·C12=18
1 3 1 C13·C33·C12=6
2 1 2 C23·C13·C22=9
1 2 2 C13·C23·C22=9

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 48×5!=5760.



Q 16 :    

Line L1 of slope 2 and line L2 of slop 12 intersect at the origin O. In the first quadrant, P1,P2,...,P12 are 12 points on line L1 and Q1,Q2,...,Q9 are 9 points on line L2. Then the total number of triangles, that can be formed having vertices at three of the 22 points OP1,P2,...,P12, Q1,Q2,...,Q9 is :          [2025]

  • 1080

     

  • 1026

     

  • 1134

     

  • 1188

     

(3)

Total number of triangles are

          =C19C212+C29C112+C19·C19C112

          = 594 + 432 + 108 = 1134



Q 17 :    

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 = C56×C35+C46×C45+C36×C55

                                      = 6×10+15×5+20×1

                                      = 60 + 75 + 20 = 155



Q 18 :    

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 = C312

                                                                             = 12×11×103×2×1 = 220

Number of ways to choose 3 points from 5 collinear points = C35 = 10

Since these 5 points are collinear, so they can not form any triangles.

   Number of triangles that can be formed = 220 –10 = 210



Q 19 :    

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

           = C212×C213=66×78 = 5148

 



Q 20 :    

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)

=C33·C27·C15·C26+C23·C37·C25·C16+C13·C47·C35·C06

= 1575 + 6300 + 1050 = 8925.