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21. Find the number of permutations of the letters of the word "Pre-University"
  A.  
13! / 2!2!2!
  B.  
13 / 3!2!1!
  C.  
12! / 2!2!3!
  D.  
11 / 2!3!4!
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm' ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here, there are 13 letters in Pre-University in which there are 2 e's, 2 i's, 2 r's and 7 other different letters.
p = 2, q = 2, r = 2, n = 13
Using these values in the shortcut, we get:
Required number of permutation =
13! / 2!2!2!

22. How many different words can be formed with the letters of the world 'University'; so that all the vowels come together?
  A.  61280
  B.  60480
  C.  58080
  D.  60080
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm'ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here,
Total numbers of letter = 10
Number of vowels = 4
i occurs 2 times.
Now, when 4 vowels are together, regarding the 4 vowels as one letter, we have only 6 + 1 = 7 letters.
Now thesee 7 letters can be arranged in 7! ways.
Since i occurs twice, therefore, 4 vowels can be arranged among themselves in
4! / 2!
ways.
∴ Required number of permutation = 7! x
4! / 2!

= 7! x
4 x 3 x 2! / 2!

= 7 x 6 x 5 x 4 x 3 x 2 x 1 x 4 x 3
= 60480

23. Find the number of rearrangements of the lettes of the word 'Benevolent'. How many of them end in l?
  A.  32380, 31250
  B.  302399, 30240
  C.  302259, 30120
  D.  303399, 30260
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm'ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here,
There are 10 letters in Benevolent in which there are 3 e's and 2 n's.
∴ Total no. of arrangements =
10! / 3! x 2!
= 302400.
But one word is benevolent itself.
∴ Number of re-arrangements =
10! / 3! x 2!
− 1 = 302399
2nd part:
When l is put in the end, number of remaining letters is 9 of which there are 3 e's and 2 n's.
∴ No. of words ending in l =
9! / 3! x 2!
= 30240

24. Letters of the word DIRECTOR are arranged in such a way that all the vowels come together. Find out the total number of ways for making such arrangement. (SBI PO)
  A.  2160
  B.  2020
  C.  2240
  D.  2150
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm'ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here,
Taking all vowels (IEO) as a single letter (since they come together) there are 6 letters with 2 R's.
∴ Total no. of arrangements = 3! x
6! / 2!
= 2160.
[3 vowels can be arranged in 3! ways among themselves, hence multiplied with 3!]

25. How many different letter arrangements can be made from the letters of the word RECOVER? (SBI PO)
  A.  1210
  B.  1260
  C.  1280
  D.  1320
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm'ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here,
Total no. of letters = 7
Numbers of R = 2
Numbers of E = 2
∴ Total no. of arrangements =
7! / 2! x 2!
= 1260.

26. In how many ways can 8 Indian, 4 Americans and 4 Englishmen be seated in a row so that all persons of the same nationality sit together.
  A.  3! 8! 4! 4!
  B.  3! 5! 4! 2!
  C.  5! 8! 4! 3!
  D.  3! 8! 7! 4!
     
   
View Answer

Here, all persons of the same nationality sit together, means all persons of the same nationality regarded as one person so we have only three persons.
These 3 persons can be seated in a row in 3! ways.
But 8 Indians can be arranged among themselves in 8! ways, 4 Americans can be arranged among themselves in 4! ways, 4 Englishman can be arranged among themselves in 4! ways.
∴ Total no. of arrangements = 3! x 8! x 4! x 4!

27. 4 boys and 2 girls are to be seated in a row in such a way that two girls are always together. In how many different ways can they be seated? (BSRB)
  A.  260
  B.  210
  C.  250
  D.  240
     
   
View Answer

Here, Assume the 2 given students to be together i.e one. Now there are 5 students. Possible ways of arranging them are 5! = 120
Now they (2 girls) can arrange themselves in 2! ways
∴ Total no. of ways = 5! x 2! = 120 x 2 = 240

28. In how many different ways can the letters of the word JUDGE be arranged so that the vowels always come together?
  A.  50
  B.  55
  C.  48
  D.  38
     
   
View Answer

Shortcut:
(I) To find the number of permutations of 'n' things taking all at a time when 'p' of them are similar and of one type, 'q' of them are similar and are of another type, 'r' of them are similar and are of third type and the remaining [n − (p + q + r)]] are all different.
The required number of permutations =
n! / p!q!r!

(II) If a work X can be done in 'n' ways and another work Y can be done in 'm'ways and Q is the final work which is done only when both X and Y are done, then the number of ways of doing the final work Q = n x m.
Here,
Taking all vowels (EU) as a single letter (since they come together) there are 4 letters.
∴ Total no. of arrangements = 4! x 2!
= 4 x 3 x 2 x 1 x 2 x 1 = 48
[2 vowels can be arranged in 2! ways among themselves, hence multiplied with 2!]

29. Find the value of n, if nC6: n-3C3 = 33 : 4
  A.  10
  B.  11
  C.  15
  D.  10
     
   
View Answer

Problems based on direct application of the following formulas
(i) nCr =
n! / r!(n − r)!


(ii) nCr − 1 + nCr = n + 1Cr

(iii) If nCr = nCq, then either r = q or r + q = n


Here, nC6 =
n! / 6!(n − 6)!

n − 3C3 =
(n − 3)! / 3!(n − 3 − 3)!
=
(n − 3)! / 3! x (n − 6)!


Hence,
n! / 6!(n − 6)!
x
3! x (n − 6)! / (n − 3)!
=
33 / 4

n! / (n − 3)!
x
3! / 6!
=
33 / 4

n(n − 1)(n − 2)(n − 3)! / (n − 3)!
x
3! / 6 x 5 x 4 x 3!
=
33 / 4

n(n − 1)(n − 2) / 6 x 5 x 4
=
33 / 4

n(n − 1)(n − 2) = 6 x 5 x 33
n(n − 1)(n − 2) = 11 x 3 x 3 x 2 x 5
n(n − 1)(n − 2) = 11 x 10 x 9
n = 11

30. Find the number of ways in which 5 identical balls can be distributed among 10 identical boxes, if not more than one ball can go into a box.
  A.  252
  B.  245
  C.  260
  D.  255
     
   
View Answer

Problems based on number of combinations.
(i) In simple cases.
(ii) When certain things are included or excluded.
Here, Number of boxes = 10
Number of balls = 5
Now distributing 5 balls among 10 boxes, when not more than one ball can go into a box amounts to selecting boxes from among the 10 boxes. This can be done in 10C5 ways
∴ Required number of ways = 10C5
=
10! / 5!(10 − 5)!

=
10! / 5! x 5!

=
10 x 9 x 8 x 7 x 6 x 5! / 5 x 4 x 3 x 2 x 1 x 5!

=
10 x 9 x 8 x 7 x 6 / 5 x 4 x 3 x 2 x 1
= 9 x 4 x 7 = 252

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