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11.
How many diagonals are there in a polygon of n sides?
  • A.
    n (n -3)
  • C.
    1  n (n -3)  
    2  
  • B.
    1  n (n -3)  
    3  
  • D.
    1  n (n -3)  
    2  
  • Answer & Explanation
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Answer : [D]
Explanation :
A polygon of n sides has n vertices.
By joining any two of these vertices, we obtain either a side or a diagonal of the polygon.
Number of all straight lines = nC2 =
n(n-1)
2
     
Number of diagonals =
n(n-1)
2
 - n =
1
2
 n (n - 3)
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12.
The number of ways in which 4 distinct balls can be put in 4 boxes labeled A, B, C, D, so that one box remains empty is
  • A.
    232
  • C.
    196
  • B.
    192
  • D.
    144
  • Answer & Explanation
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Answer : [D]
Explanation :
Let the balls be named a,b,c,d and the four boxes be A,B,C and D.
One box should always remain empty.
Let us assume initially, that the box A remains empty.
Now there are 4 balls to be put in 3 Boxes B, C, D.
Hence one of the 3 boxes should receive 2 balls. Let this box be B.
The two balls to be put in B can be chosen in 4C2 = 6 ways.
The box C can be filled with 1 ball out of 2 balls in 2C1 = 2 ways.
Hence if A is kept empty and B is filled with 2 balls and C and D are filled with 1 ball in
4C2 x 2 = 12 ways.
Similarly, we can put 2 balls in box C or 2 balls in box D, Keeping A empty.
Hence it can be done in 12 x 3 = 36 ways.
Since any of the 4 boxes, A, B, C and D could be empty we have in all 36 x 4 = 144 ways.
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13.
There are P points in a plane of which only K points are in a straight line. Find the umber of triangles which can be formed by the P points?
  • A.
    1  [P(P-1)(P-2) - K(K - 1)(K - 2)]  
    6  
  • C.
    PC3 - K
  • B.
    P! - K!
  • D.
    P! - KC3
  • Answer & Explanation
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Answer : [A]
Explanation :
The number of triangles = the number of ways in which 3 points can be chosen out of P points the number of ways in which 3 points can be chosen out of K points.
= PC3 - KC3 =
P!
3!(P - 3)!
-
K!
3!(K - 3)!
       
=
1
6
 [P(P -1)(P-2) - K(K -1) (K - 2)]      
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