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- Permutation and Combination
11.
How many diagonals are there in a polygon of n sides?
- A.n (n -3)
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C.
1 n (n -3) 2
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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. | |||||||||
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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
- 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 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. |
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?
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A.
1 [P(P-1)(P-2) - K(K - 1)(K - 2)] 6 - C.PC3 - K
- B.P! - K!
- D.P! - KC3
- 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. | ||||||||||||
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