Home / Aptitude / Permutation / Questions based on Digits and Numbers
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Q1. |
How many 9 digit numbers are there in which all the digits are distinct? |
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A. 9 x 9! |
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B. 9 x 8! |
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C. 9! |
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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? |
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A. 125 |
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B. 500 |
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C. 96 |
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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? |
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A. 256 |
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B. 64 |
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C. 192 |
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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? |
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A. 180 |
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B. 216 |
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C. 360 |
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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? |
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A. 1680 |
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B. 6720 |
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C. 3360 |
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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? |
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A. 41 |
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B. 32 |
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C. 44 |
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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 one-digit 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? |
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A. 48 |
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B. 24 |
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C. 18 |
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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? |
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A. 24 |
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B. 48 |
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C. 96 |
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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? |
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A. 864 |
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B. 863 |
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C. 648 |
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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? |
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A. 36 |
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B. 48 |
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C. 30 |
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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? |
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A. 48 |
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B. 57 |
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C. 96 |
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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? |
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A. 313 |
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B. 229 |
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C. 192 |
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D. 193 |
From the digits 4, 5, 6, and 7 means repetition is allowed.
One digit number: 4
Two-digit numbers: 4 × 4 =16
Three-digit numbers: 4 × 4 × 4 = 64
Four-digit numbers: beginning with 4: 4 × 4 × 4 = 64
Four-digit numbers: beginning with 5: 4 × 4 × 4 = 64
Four-digit numbers: beginning with 6: 4 × 4 × 4 = 64
Four-digit numbers: beginning with 74: 4 × 4 = 16
Four-digit numbers: beginning with 75: 4 × 4 = 16
Four-digit numbers: beginning with 764: = 4
Four-digit 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? |
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A. 96 |
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B. 144 |
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C. 108 |
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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