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11. A letter lock consists of three rings each marked with 10 different letters. In how many ways it is possible to make an unsuccessful attempt to open the lock?
  A.  666
  B.  777
  C.  888
  D.  999
     
   
View Answer

Shortcut:
Number of permutations when repetition is allowed

Number of permutations of 'n' different things taken 'r' at a time when things can be repeated any number of times = n x n x n ........r times = nr.

[Note: In such type of problems, you've to first determine as to which item can be repeated. And consider the value of repeated item as 'r' in the above formula]

Here, two rings may have same letter at a time but same ring cannot have two letters at a time, therefore, we must start with ring.

Each of the three rings can have any one of the 10 different letters in 10 ways.

∴ Total number of attempts = 10 x 10 x 10 = 103 = 1000

But out of these 1000 attempts only one attempt is successful.

∴ Required number of unsuccessful attempts = 1000 − 1 = 999

12. In a class of 10 students, there are 3 girls. In how many different ways can they be arranged in a row such that no two of the three girls are consecutive?
  A.  
7! / 8!
×5!
  B.  
8! / 7!
×5!
  C.  
8! / 5!
×7!
  D.  
5! / 8!
×7!
     
   
View Answer

Shortcut:
If there are two groups X and Y consisting of 'a' and 'b' things respectively, then the number of ways in which no two of group Y occur together are given by a + 1Pb x a!.

Provided that b <= a.

Here, a = 7, b = 3
Using these values in the shortcut, we get:
= 7 + 1P3 x 7!.
= 8P3 x 7!.
=
8! / (8 − 3)!
x 7! =
8! / 5!
x 7!

13. In how many ways can 7 B. A and 5 B.Sc students be seated in a row so that no two of B. Sc students may sit together?
  A.  
8! / 5
×3!
  B.  
8! / 3!
×7!
  C.  
8! / 3!
! ×6!
  D.  
3! / 8!
×7!
     
   
View Answer

Shortcut:
If there are two groups X and Y consisting of 'a' and 'b' things respectively, then the number of ways in which no two of group Y occur together are given by a + 1Pb x a!.

Provided that b <= a.

Here, a = 7, b = 5
Using these values in the shortcut, we get:
= 7 + 1P5 x 7!.
= 8P5 x 7!.
=
8! / (8 − 5)!
x 7! =
8! / 3!
x 7!

14. In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together.
  A.  8 x 9!
  B.  7 x 9!
  C.  9 x 7!
  D.  9 x 8!
     
   
View Answer

Shortcut:
The number of ways in which 'm' examination papers can be arranged so that the best and the worst papers never come together are given by [(m − 2) x (m − 1)!] ways.
Here, m = 10
Using these values in the shortcut, we get:
= [(10 − 2) x (10 − 1)!] = 8 x 9!

15. There are 5 boys and 3 girls. In how many ways can they be seated in a row so that all the three girls do not sit together.
  A.  32000
  B.  38000
  C.  34000
  D.  36000
     
   
View Answer

Shortcut:
There are 'a' boys and 'b' girls. The number of ways in which they can be seated in a row so that all the girls do not sit together are given by [(a + b)! − (a + 1)! x b!] ways.
Here, a = 5, b = 3
Using these values in the shortcut, we get:
= [(5 + 3)! − (5 + 1)! x 3!]
= 8! − 6! x 3!
= 6!(8 x 7 − 3!)
= 6!(56 − 3 x 2 x 1)
= 6!(56 − 6)
= 6! x 50
= 6! x 50
= 36000

16. In how many ways 4 boys and 4 girls can be seated in a row so that boys and girls are alternate?
  A.  1142
  B.  1152
  C.  1148
  D.  1154
     
   
View Answer

Shortcut:
The number of ways in which a boys and 'a' girls can be seated in a row so that boys and girls are alternate are given by 2(a!a!) ways.
Here, a = 4
Using these values in the shortcut, we get:
= 2(a!a!)
= 2(4!4!)
= 2(4 x 3 x 2 x 1 x 4 x 3 x 2 x 1)
= 2(24 x 24)
= 2 x 576 = 1152

17. In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate?
  A.  100
  B.  169
  C.  144
  D.  121
     
   
View Answer

Shortcut:
The number of ways in which a boys and '(a − 1)' girls can be seated in a row so that they are alternate are given by [a!(a − 1)!] ways.
Here, a = 4, (a − 1) = 3
Using these values in the shortcut, we get:
= [4!3!]
= [4 x 3 x 2 x 1 x 3 x 2 x 1]
= 24 x 6 = 144

18. In how many ways can 5 Indians and 5 English men be seated along a circle so that they are alternate?
  A.  4! 5!
  B.  3! 6!
  C.  4! 6!
  D.  3! 5!
     
   
View Answer

Shortcut:
The number of ways in which 'a' persons of a particular group,, caste , country etc and 'a' persons of the other group, caste, community, country etc can be seated along a circle so that they are alternate, given by [a!(a − 1)!] ways.
Here, a = 5, a = 5
Using these values in the shortcut, we get:
= [5!(5 − 1)!]
= 5! x 4!

19. A round table conference is to be held between 20 delegates of 20 countries. In how many ways can they be seated if two particular delegates are always to sit together?
  A.  16! × 2!
  B.  18! × 2!
  C.  18! × 3!
  D.  19! × 2!
     
   
View Answer

Shortcut:
A round table conference is to be held between 'a' delegates. The number of ways in which they can be seated so that 'b' particular delegates always sit together re given by [(a − b)! x b!] ways.
Here, a = 20, b = 2
Using these values in the shortcut, we get:
= [(20 − 2)! x 2!]
= [18! x 2!]

20. Find the number of ways in which 6 different beads can be arranged to form a necklace.
  A.  72 ways
  B.  66 ways
  C.  60 ways
  D.  54 ways
     
   
View Answer

Shortcut:
The number of ways in which 'a' different beads can be arranged to form a necklace are given by
(a − 1)! / 2

Here, a = 6
Using these values in the shortcut, we get:
=
(6 − 1)! / 2
=
5! / 2

=
5 x 4 x 3 x 2 x 1 / 2
= 5 x 4 x 3 = 60
Hence, 6 different beads can be arranged to form a necklace is 60 ways.

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