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51. In how many ways 20 apples can be divided among 5 boys.
  A.  10626
  B.  10428
  C.  11228
  D.  10426
     
   
View Answer

Shortcut:
The number of ways in which 'm' identical things can be divided or distributed among 'z' persons when any person may get any number of things are m + z − 1Cz − 1
Here, m = 20, z = 5
Using the value in the shortcut, we get:
Required ways = 20 + 5 − 1C5 − 1
= 24C4
=
24! / (4!)(24 − 4)!

=
24! / 4! x 20!

=
24 x 23 x 22 x 21 x 20! / 4 x 3 x 2 x 1 x 20!

= 23 x 22 x 21 = 10626
Hence, we can divide in 10626 ways.

52. Find the number of quadrilaterls that can be formed by joining the vertices of an octagon.
  A.  85
  B.  70
  C.  90
  D.  55
     
   
View Answer

Shortcut:
The number of quadrilaterals that can be formed by joining the vertices of a polygon of 'm' sides are given by
m(m − 1)(m − 2)(m − 3) / 24
; where m > 3.
Here, m = 8 [As octagon has 8 sides]
Using the value in the shortcut, we get:
Required ways =
8(8 − 1)(8 − 2)(8 − 3) / 24

=
8 x 7 x 6 x 5 / 24

= 7 x 2 x 5 = 70
Hence, we can form 70 quadrilaterls.

53. How many straight lines can be drawn with 16 points in a plane of which no points are collinear?
  A.  125
  B.  110
  C.  120
  D.  135
     
   
View Answer

Shortcut:
If there are 'm' points in a plane and no points are collinear, then the number of straight lines can be drawn using these 'm' points are given by
m(m − 1) / 2
.
Here, m = 16
Using the value in the shortcut, we get:
Required ways =
16(16 − 1) / 2

=
16 x 15 / 2

= 8 x 15 = 120
Hence, we can draw 120 straight lines.

54. Find the number of triangles that can be formed with 12 points in a plane of which no three points are collinear.
  A.  222
  B.  216
  C.  220
  D.  210
     
   
View Answer

Shortcut:
If there are 'm' points in a plane and no three points are collinear, then the number of triangels formed with 'm' points are given by
m(m − 1)(m − 2) / 6
.
Here, m = 12
Using the value in the shortcut, we get:
Required ways =
12(12 − 1)(12 − 2) / 6

=
12 x 11 x 10 / 6

= 2 x 11 x 10 = 220
Hence, we can form 220 triangles.

55. 6 students appear in an examination. In how many ways can the result be announced?
  A.  64
  B.  68
  C.  58
  D.  60
     
   
View Answer

Shortcut:
'm' students appear in an examination. The number of ways the result of the examination can be announced are given by (2)m.
Here, m = 6
Using the value in the shortcut, we get:
Required ways = (2)6
= 2 x 2 x 2 x 2 x 2 x 2 = 64
Hence, we can announce result in 64 ways.

56. 3 matches are to be played in a chess tournament. In how many ways can their results be decided?
  A.  33
  B.  21
  C.  25
  D.  27
     
   
View Answer

Shortcut:
'm' matches are to be played in a chess tournament. The number of ways in which their results can be decided are given by (3)m.
Here, m = 3
Using the value in the shortcut, we get:
Required ways = (3)3
= 3 x 3 x 3 = 27
Hence, we can announce result in 27 ways.

57. From a group of 6 men and 4 women a committee of 4 persons is to formed.
(i) In how many different ways can it be done so that the committee has at least one woman?
(ii) In how many different ways can it be done so that the committee has at least 2 men? (SBI)
  A.  (i): 195; (ii): 185
  B.  (i): 175; (ii): 165
  C.  (i): 185; (ii): 180
  D.  (i): 180; (ii): 160
     
   
View Answer

(i) 4C1 x 6C3 + 4C2 x 6C2 + 4C3 x 6C1 + 4C4
=
4! / (1!)(4 − 1)!
x
6! / (3!)(6 − 3)!
+
4! / (2!)(4 − 2)!
x
6! / (2!)(6 − 2)!
+
4! / (3!)(4 − 3)!
x
6! / (1!)(6 − 1)!
+
4! / (4!)(4 − 4)!

=
4! / 3!
x
6! / 3! x 3!
+
4! / 2! x 2!
x
6! / 2! x 4!
+
4! / 3! x 1!
x
6! / 1! x 5!
+
4! / (4!)0

= 4 x
6 x 5 x 4 / 3 x 2 x 1
+
4 x 3 / 2 x 1
x
6 x 5 / 2 x 1
+ 4 x 6 + 1
= 4 x 5 x 4 + 2 x 3 x 3 x 5 + 24 + 1
= 80 + 90 + 24 + 1 = 195
Hence, we can form in 195 ways.

(ii) 6C2 x 4C2 + 6C3 x 4C1 + 6C4
=
6! / (2!)(6 − 2)!
x
4! / (2!)(4 − 2)!
+
6! / (3!)(6 − 3)!
x
4! / (1!)(4 − 1)!
+
6! / (4!)(6 − 4)!

=
6! / 2! x 4!
x
4! / 2! x 2!
+
6! / 3! x 3!
x
4! / 1! x 3!
+
6! / 4! x 2!

=
6 x 5 x 4! / 2x 1 x 4!
x
4 x 3 x 2! / 2x 1 x 2!
+
6 x 5 x 4 x 3! / 3 x 2 x 1 x 3!
x
4 x 3! / 1! x 3!
+
6 x 5 x 4! / 4! x 2 x 1

= 3 x 5 x 3 x 2 + 5 x 4 x 4 + 3 x 5
= 90 + 80 + 15 = 185

58. From 4 officers and 8 jawans in how many ways can 6 be chosen to include at least one officer?
  A.  896
  B.  860
  C.  880
  D.  986
     
   
View Answer

Shortcut:
From 'a' persons of a group X and 'b' persons from group Y, the number of ways in which 'n' persons can be chosen to include exactly 'p' persons of group X and the rest of group Y is given by (aCp x bCn − p) ways
Here, a = 4, b = 8, n = 6, p = 1
Using the value in the shortcut, we get:
Required ways = 4C1 x 8C5 + 4C2 x 8C4 + 4C3 x 8C3 + 4C4 x 8C2
=
4! / 1!(4 − 1)!
x
8! / 5! x (8 − 5)!
+
4! / 2!(4 − 2)!
x
8! / 4! x (8 − 4)!
+
4! / 3!(4 − 3)!
x
8! / 3! x (8 − 3)!
+
4! / 4!(4 − 4)!
x
8! / 2! x (8 − 2)!

=
4! / 3!
x
8! / 5! x 3!
+
4! / 2! x 2!
x
8! / 4! x 4!
+
4! / 3! x 1!
x
8! / 3! x 5!
+
4! / 4! x 0
x
8! / 2! x 6!

= 224 + 420 + 224 + 28 = 896
Hence, we can choose in 896 ways.

59. A badminton tournament consists of 3 matches.
(i) In how many ways can their results be forecast.
(ii) How many different forecasts cna contain all wrong results.
(iii) How many different forecasts can contain all correct results?
  A.  (i):8; (ii): 3; (iii): 5
  B.  (i):8; (ii): 1; (iii): 1
  C.  (i):6; (ii): 5; (iii): 7
  D.  (i):9; (ii): 3; (iii): 4
     
   
View Answer

Shortcut:
A badminton tournament consists of 'm' matches.
(a) The number of ways in which their results can be forecast are given by (2)m.
(b) Total number of forecasts containing all correct results or all wrong results are given by 1.
Here, m = 3
Using the value in the shortcut, we get:
(i) Each badminton match can be decided in only 2 ways i.e win or loss for a particular team.
∴ Total number of ways the results of 3 matches can be forecast = (2)3 = 8

(ii) Result of each match can be forecast wrong in only 1 way.
∴ Totalnumber of forecasts containing all wrong results = (1)3 = 1

(iii) Similarly, result of each match can be forecast correct in only 1 way.
∴ Total number of forecasts containing all correct results = (1)3 = 1

60. The Indian hockey team is to play 4 matches for the world cup.
(i) How many different forecasts will contain all correct results?
(ii) How many different forecasts will contain all wrong results?
  A.  (i):16; (ii): 1;
  B.  (i):3; (ii): 20;
  C.  (i):3; (ii): 12;
  D.  (i):1; (ii): 16;
     
   
View Answer

Shortcut:
The Indian hockey team is to play 'm' matches for the world cup. (a) The number of different forecasts that will contain all correct results is (1)m or 1.
(b) The number of different forecasts that will contain all wrong results are (2)m.
Here, m = 4
Using the value in the shortcut, we get:
Result of each hockey match can be decided in 3 ways i.e win, loss or draw.
(i) Only in 1 way we can correctly predict each match i.e 'forecast' and the actual result are the same.
Hence, total number of ways to predict all 4 matches correctly = (1)4 = 1

(ii) For each match the prediction can go wrong in 2 ways. For example, prediction for any match is win but the actual result i.e. either loss or draw.
∴ Total number of forecasts containing all wrong results = (2)4 = 16.

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