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 General Aptitude
 Permutation and Combination
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
 Report
Answer : [D]
Explanation :
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.  




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
 Report
Answer : [D]
Explanation :
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 4C_{2} = 6 ways. 
The box C can be filled with 1 ball out of 2 balls in 2C_{1} = 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 
4C_{2} 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. 
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(P1)(P2)  K(K  1)(K  2)] 6  C.PC_{3}  K
 B.P!  K!
 D.P!  KC_{3}
 Answer & Explanation
 Report
Answer : [A]
Explanation :
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.  



