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Q1. 
How many 9 digit numbers are there in which all the digits are distinct? 
A. 9 x 9! 

B. 9 x 8! 

C. 9! 

D. 8! 
First place can be filled in 9 ways
Hence total number = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 × 9 !
Q2. 
How many 5 digit numbers can be formed with the digits 4, 5, 6, 7, and 0? 
A. 125 

B. 500 

C. 96 

D. 48 
4 × 4 × 3 × 2 × 1 = 96
Repetition is not allowed.
Q3. 
How many 4 digit numbers can be formed using the digits 1, 2, 3, 4? 
A. 256 

B. 64 

C. 192 

D. 24 
Note: whenever using/with the digits/given the digits is asked, repetition is not allowed.
Ans: 4 × 3 × 2 × 1 = 24
Q4. 
How many 4 digit numbers can be formed from the digits 3, 5, 6, 7, and 0? 
Q5. 
How many 4 digit numbers can be formed from the digits 2, 5, 6, 7, and 0 if repetition of digits is not allowed? 
Q6. 
How many 3 digit numbers can be formed from the digits 2, 4, 5, and 6 if repetition of digits is not allowed? 
Q1. 
How many numbers greater than four million can be formed with the digits 2, 3, 0, 3, 4, 2 and 5? 
A. 180 

B. 216 

C. 360 

D. 720 
Greater than four million will be formed with the digits 2, 3, 0, 3, 4, 2 and 5.
The first place will be occupied by either 4 or 5.
The remaining places will be occupied by 6 !/ 2! × 2! = 180
So Total number of ways = 2 × 180 = 360
Q2. 
How many 7 digit telephone numbers can be constructed if each number starts with 91 and no digit appears more than once? 
A. 1680 

B. 6720 

C. 3360 

D. 13440 
If each number starts with 91,
Remaining digits left = 8 (Out of 10 digits)
And remaining places left = 5
So total numbers = 8 × 7 × 6 × 5 × 4 = 6720
Q3. 
How many even numbers less than 10,000 can be formed with the digits from the digits 1, 3, 4, 5, and 7? 
A. 41 

B. 32 

C. 44 

D. 16 
1, 3, 4, 5, and 7
Less than 10, 000 means there can be 1 digit, 2 digits, 3 digits, and 4 digits numbers.
There is only 1 even onedigit number i.e. 4
2 digits: 4 × 1 = 4
3 digits: 4 × 3 × 1 = 12
4 digits: 4 × 3 × 2 × 1 = 24
Total = 1 + 4 + 12 + 24 = 41
Q4. 
How many 4 digit even numbers can be formed with distinct digits from the digits 1, 3, 4, 5 and 0? 
A. 48 

B. 24 

C. 18 

D. 42 
1, 3, 4, 5 and 0
Keeping '0' in the units digit: 4 × 3 × 2 × 1 = 24
Keeping '4' in the units digit: 3 × 3 × 2 × 1 = 18
Total = 24 + 18 = 42 numbers
Q5. 
How many 4 digit odd numbers can be formed with distinct digits from the digits 2, 3, 4, 5 and 8? 
A. 24 

B. 48 

C. 96 

D. 12 
The last will be filled in 2 ways i.e. 3 or 5.
Total number = 4 × 3 × 2 × 2 = 48
Q6. 
How many 4 digit numbers from 5000 to 8000 can be formed from the digits 4,5,6,7,8 and 0? 
A. 864 

B. 863 

C. 648 

D. 649 
How many 4 digit numbers from 5000 to 8000 can be formed from the digits 4,5,6,7,8 and 0?
The thousand placed can be filled in 3 ways i.e. 5 or 6 or 7
The other places can be filled in 6 ways.
3 × 6 × 6 × 6 = 648
As 8000 will be included, the answer would be 648 + 1= 649
Q1. 
How many 5 digit numbers s which are divisible by 4 can be formed using the digits 1, 2, 3, 4, and 0? 
A. 36 

B. 48 

C. 30 

D. 24 
Repetition is not allowed as we have to use all the digits
To be divisible by 4, the last 2 digits must be: 12, 04, 20, 24, 32, 40
Keeping 04 or 20 or 40 in the last two places, numbers = 3 × 2 × 1 × 3 = 18
Keeping 12 or 24 or 32 in the last two places, numbers = 2 × 2 × 1 × 3 = 12
Q2. 
How many even numbers with distinct digits greater than 500 can be formed from the digits 3, 4, 5, 6, and 7? 
A. 48 

B. 57 

C. 96 

D. 111 
Even numbers greater than 500 to be formed with the digits 3, 4, 5, 6 and 7.
3 digit numbers :
Keeping 4 in units place : 3 × 3 × 1 =9 (5,6 and 7 in the Hundreds place)
Keeping 6 in units place: 2 × 3 × 1 =6 (5 & 7 in the Hundreds place)
4 digit numbers :
Keeping 4 or 6 in units place : 4 × 3 × 2 × 2 = 48
5 digit numbers :
Keeping 4 or 6 in units place: 4 × 3 × 2 × 1 × 2 = 48
Total numbers = 9 + 6 + 48 + 48 = 111
Q3. 
How many numbers not exceeding 7654 can be formed from the digits 4, 5, 6, and 7? 
A. 313 

B. 229 

C. 192 

D. 193 
From the digits 4, 5, 6, and 7 means repetition is allowed.
One digit number: 4
Twodigit numbers: 4 × 4 =16
Threedigit numbers: 4 × 4 × 4 = 64
Fourdigit numbers: beginning with 4: 4 × 4 × 4 = 64
Fourdigit numbers: beginning with 5: 4 × 4 × 4 = 64
Fourdigit numbers: beginning with 6: 4 × 4 × 4 = 64
Fourdigit numbers: beginning with 74: 4 × 4 = 16
Fourdigit numbers: beginning with 75: 4 × 4 = 16
Fourdigit numbers: beginning with 764: = 4
Fourdigit numbers: beginning with 765: = 1 i.e= (7654)
Total numbers = 4+16+64+64+64+64+16+16+1 = 313
Q4. 
How many 5 digit numbers divisible by 6 can be formed from the digits 1, 2, 3, 4, 5, and 0 if repetition of digits is not allowed? 
A. 96 

B. 144 

C. 108 

D. 72 
To be divisible by 6, the numbers must be divisible by 2 & 3.
Case 1: Excluding '0' , we have 5 digits 1,2,3,4,& 5 whose sum = 15
This concludes any 5 digit numbers formed with 1,2,3,4and 5 will be divisible by 3.
Again the numbers must be even to be divisible by 2.
Numbers formed with the condition = 4 × 3 × 2 × 1 × 2 = 48
(Last digits will be occupied by either 2 or 4)
Case 2: Including 0 and Excluding '3' , we have 5 digits 1,2,4,5. 0 whose sum = 12
This concludes any 5 digit numbers formed with 1, 2,4,5, and 0 will be divisible by 3.
Again the numbers must be even to be divisible by 2.
Last digits will be occupied by either 0,2, or 4
Keeping 0 in the end: 4 × 3 × 2 × 1 × 1 = 24
Keeping 2/4 in the end: 3 × 3 × 2 × 1 × 2 = 36
Total numbers formed = 48 + 24 + 36 = 108