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1. If n P4 = 360, find the value of n.
  A.  6
  B.  4
  C.  2
  D.  8
     
   
View Answer

Application of the direct formula of permutation.
nPr =
n! / (n − r)!
∴ nP4 =
n! / (n − 4)!

n! / (n − 4)!
= 360
or,
n(n − 1)(n − 2)(n − 3)(n − 4)! / (n − 4)!
= 360
or, n(n − 1)(n − 2)(n − 3) = 360
or, n(n − 1)(n − 2)(n − 3) = 6 x 5 x 4 x 3
∴ n = 6
[Note: Here LHS is the product of four consecutive integers, therefore, we have to express 360 into product of four consecutive integers.

360 = 2 x 2 x 2 x 3 x 3 x 5, greatest of these factors is 5, therefore try with 5.

Integers just before and after 5 are 4 and 6. Both 4 and 6 are factors of 360. Thus we get four consecutive integers 6, 5, 4 and 3 whose product is 360.

If 5 does not suit, then try with 2 x 5 i.e 10 etc.]

2. If 10 Pr = 720, find the value of r.
  A.  8
  B.  3
  C.  9
  D.  2
     
   
View Answer

Application of the direct formula of permutation.
nPr =
n! / (n − r)!

∴ 10Pr =
10! / (10 − r)!

10! / (10 − r)!
= 720
or, 10 x 9 x 8 x ......... to r factors = 720
or, 10 x 9 x 8 x ......... to r factors = 10 x 9 x 8
∴ r = 3

3. How many numbers of four digits can be formed with the digits 1, 2, 3, 4 and 5? (If repetition of digits is not allowed)
  A.  110
  B.  115
  C.  120
  D.  125
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.
First of all decide of how many digits the required numbers will have.
Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.
Thousands place Hundreds place Tens place Units place

Here n = Number of digits = 5
and r = number of places to be filled up, i.e 4
       
∴ Required number = 5P4 =
5! / (5 − 4)!

∴
5! / 1!
= 5 x 4 x 3 x 2 x 1 = 120

4. How many numbers between 400 and 1000 can be made with the digits 2, 3, 4, 5, 6 and 0?
  A.  40
  B.  50
  C.  60
  D.  70
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.

First of all decide of how many digits the required numbers will have.

Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

Thousands place Hundreds place Tens place Units place

Here nothing has been given about repetition of digits. Therefore, we will assuem that repetition of digits is not allowed.
Any number between 400 and 1000 must be of three digits only.

4 or 5 or 6    

Since the number should be greater than 40, therefore, hundreds place can be filled up by any one of the three digits 4,5 and 6 in 3 ways. Remaining two places can be filled up by remaining five digits in 5P2 ways.
∴ Required number = 3 x 5P2 = 3 x
5! / (5 − 2)!

∴ 3 x
5! / 3!
= 3 x
5 x 4 x 3! / 3!
= 3 x 5 x 4 = 60

5. Find the number between 300 and 3000 that can be formed with the digits 0, 1, 2, 3, 4, and 5, no digits being repeated in any number.
  A.  180
  B.  160
  C.  170
  D.  190
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.

First of all decide of how many digits the required numbers will have.

Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

Thousands place Hundreds place Tens place Units place

Here any number between 300 and 3000 must be of 3 or 4 digits.

Case I: When number is of 3 digits.
3 or 4 or 5    

Hundreds place can be filled up by any one of the three digits 3 4 and 5 in 3 ways.

Remaining two places can be filled up remaining five digits in 5P2 ways.

∴ Required number = 3 x 5P2 = 3 x
5! / (5 − 2)!

∴ 3 x
5! / 3!
= 3 x
5 x 4 x 3! / 3!
= 3 x 5 x 4 = 60
Case II: When number is of 4 digits.
1 or 2      
Thousands place can be filled up by any one of the two digits 1 and 2 in 2 ways and the remaining three places can be filled up by remaining five digits in 5P3 ways.
∴ Numbers formed in this case = 2 x 5P3 = 2 x
5! / (5 − 3)!

= 2 x
5! / 2!

= 2 x
5 x 4 x 3 x 2! / 2!
= 2 x 5 x 4 x 3 = 120
Hence, the required number = 60 + 120 = 180

6. How many even numbers of four digits can be formed with the digits 0, 1, 2, 3, 4, 5 and 6? (No digit is repeated more than once)
  A.  400
  B.  410
  C.  430
  D.  420
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.

First of all decide of how many digits the required numbers will have.

Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

Thousands place Hundreds place Tens place Units place

Here Each even number must have 0m 2, 4 or 6 in its units' place. Here total number of digits = 7.
      0 or 2 or 4 or 6

[When 0 occurs at units place there is no restriction on other places and when 2 or 4 or 6 occurs at units place there is restriction on thousands place as 0 can not be put at thousands' place]
Case I: When 0 occurs at units' place.
      0
Unit's place can be filled up 0 in 1 way and remaining three places can be filled up by remaining 6 digits in 6P3 ways.
∴ Number of formed in this case = 1 x 6P3
= 1 x
6! / (6 − 3)!

∴ 1 x
6! / 3!
= 1 x
6 x 5 x 4 x 3! / 3!
= 1 x 6 x 5 x 4 = 120
Case II: When 0 does not occur at units' place.
any one of remaining six digits except zero.     2 or 4 or 6
Units' place can be filled up by any one of the three digits 2, 4 and 6 in 3 ways. Thousands' place can be filled up by any one of the remaining six digits except zero in 5 ways. Remaining two places can be fille dup by remaining five digits in 5P2 ways. ∴ Numbers of numbers of formed in this case = 5 x 3 x 5P2
= 5 x 3 x
5! / (5 − 2)!

∴ 15 x
5! / 3!
= 15 x
5 x 4 x 3! / 3!
= 15 x 5 x 4 = 300
∴ Required number = 120 + 300 = 420

7. How many numbers of four digits greater than 2300 can be formed with the digits 0, 1, 2, 3, 4, 5 and 6? (No digit being repeated)
  A.  560
  B.  550
  C.  565
  D.  555
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.

First of all decide of how many digits the required numbers will have.

Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

Thousands place Hundreds place Tens place Units place

Here, Since number must be of four digits and greater than 2300, therefore any one of the five digits 2, 3, 4, 5 and 6 will occur at thousands' place. When any one of 3, 4, 5 and 6 occurs at thousands' place the number will be definitely greater than 2300 but when 2 occurs at thousands' place there will be also restriction on hundreds' place to make the number greater than 2300.

Case I: When 2 occurs at thousands' place:
2 3 or 4 or 5 or 6    

[Thousands' place can be filled up by 2 in one way and hundreds' place can be filled up by any one of the four digits 3, 4, 5 and 6 in four ways.

Remaining two places can be filled up by remaining five digits in 5P2 ways.
∴ Number of numbers formed in this case = 1 x 4 x 5P2
= 4 x
5! / (5 − 2)!

= 4 x
5! / 3!
= 4 x
5 x 4 x 3! / 3!
= 4 x 5 x 4 = 80
Case II: When any one of 3, 4, 5 and 6 occurs at thousands' place.
3 or 4 or 5 or 6      
Thousands' place can be filled up by any one of the four digits 3, 4, 5 and 6 in four ways and remaining three places can be filled up by remaining six digits in 6P3 ways.

∴ Numbers of numbers of formed in this case = 4 x 6P3
= 4 x
6! / (6 − 3)!

∴ 4 x
6! / 3!
= 4 x
6 x 5 x 4 x 3! / 3!
= 4 x 6 x 5 x 4 = 480
∴ Required number = 80 + 480 = 560

8. How many positive numbers can be formed by using any number of the digits 0, 1, 2, 3, 4, 5 and 6? (No digit is repeated)
  A.  260
  B.  250
  C.  265
  D.  245
     
   
View Answer

Shortcut:
Formation of numbers with digits when repetition of digits is not allowed.

First of all decide of how many digits the required numbers will have.

Then fill up the places on which there are restrictions and then apply the formula nPr for filling up the remaining places with remaining digits.

Thousands place Hundreds place Tens place Units place

Here, Case I: When number is of 5 digits:
1 or 2 or 3 or 4        

[Ten thousands' place can be filled up by any one of the four digits 1, 2, 3 and 4 in 4 ways and the remaining four places can be filled up by the remaining four digits in 4P4 ways.
∴ Number of numbers formed in this case = 4 x 4P4
= 4 x
4! / (4 − 4)!
= 4 x 4 x 3 x 2 x 1 = 96
Case II: When number is of four digits.
1 or 2 or 3 or 4      
Thousands' place can be filled up by any one of the four digits 1, 2, 3 and 4 in 4 ways and the remaining 3 places can be filled up by the remaining 3 digits in 4P3 ways.
∴ Numbers of numbers of formed in this case = 4 x 4P3
= 4 x
4! / (4 − 3)!

∴ 4 x
4 x 3 x 2 x 1! / 1!
= 4 x 4 x 3 x 2 = 96
Case III: When number is of 3 digits.
1 or 2 or 3 or 4    
Hundreds' place can be filled up by any one of the four digits 1, 2, 3 and 4 in 4 ways and the remaining 2 places can be filled up by the remaining 2 digits in 4P2 ways.
∴ Numbers of numbers of formed in this case = 4 x 4P2
= 4 x
4! / (4 − 2)!

∴ 4 x
4 x 3 x 2! / 2!
= 4 x 4 x 3 = 48
Case IV: When number is of 2 digits.
1 or 2 or 3 or 4  
Tens' place can be filled up by any one of the four digits 1, 2, 3 and 4 in 4 ways and the remaining 1 place can be filled up by the remaining 1 digit in 4P1 ways.
∴ Numbers of numbers of formed in this case = 4 x 4P1
= 4 x
4! / (4 − 1)!

∴ 4 x
4 x 3! / 3!
= 4 x 4 = 16
Case V: When number is of 1 digit.
∴ Numbers of numbers of formed in this case = 4
∴ Required number = 96 + 96 + 48 + 16 + 4 = 260

9. A man has 6 friends to invite. In how many ways can he send invitation cards to them if he has three servants to carry the cards.
  A.  784
  B.  729
  C.  841
  D.  676
     
   
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]

In this question,
[1st servant or 2nd servant] ----> same friend

[1st friend or 2nd friend] ----> same servant

Here, we observe that invitation cards cannot be sent to the same friend by different servants but invitation cards may be sent to different friends by the same servant. Thus same servant may be repeated for different frineds, therefore, we must start with frined.

Invitation cards may be sent to each of the six friends by any one of the trhee servants in 3 ways.

∴ Required number = 3 x 3 x 3 x 3 x 3 x 3 = 36 = 729

10. In how many ways 3 prizes can be given away to 7 boys when each boy is eligible for any of the prizes.
  A.  125
  B.  512
  C.  216
  D.  343
     
   
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]

In this question,
[1st prize or 2nd prize] ----> same boy

[1st boy or 2nd boy] ----> same prize

Here, we observe that two prizes can be given to the same boy but two boys cannot get the same prize, therefore, we must start with prize.

Each of the three prizers can be given away to any one of the 7 boys in 7 ways.

∴ Required number = 7 x 7 x 7 = (7)3 = 343

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