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