CRE 4 - Forming numbers using digits | Modern Math - Permutation & Combination
Answer the next 10 questions based on the information given below:
Consider the digits 2, 3, 4, 7 & 8. Using these digits,
How many 5-digit numbers can be formed, if repetition of digits is not allowed?
Answer: 120
Explanation :
Let us label the digits of 5-digit number as 5 4 3 2 1.
The rightmost digits i.e., 1st digit can be filled by either of 2 or 3 or 4 or 7 or 8, i.e., in 5 ways.
Now since repetition is not allowed, 2nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.
Similarly, 3rd digit can be filled by remaining 3 numbers in 3 ways.
∴ 4th digit can be filled in 2 days.
and 5th digit can be filled in 1 way.
⇒ Total number of such 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120.
Hence, 120.
Workspace:
How many 5-digit numbers can be formed, if repetition of digits is allowed?
Answer: 3125
Explanation :
Let us label the digits of 5-digit number as 5 4 3 2 1.
The rightmost digits i.e., 1st digit can be filled by either of 2 or 3 or 4 or 7 or 8, i.e., in 5 ways.
Now since repetition is allowed, 2nd digit can also be filled by any of these 5 numbers, i.e., in 5 ways.
Similarly, 3rd, 4th and 5th digits can also be filled by any of these 5 numbers in 5 ways each.
⇒ Total number of such 5-digit numbers = 5 × 5 × 5 × 5 × 5 = 3125.
Hence, 3125.
Workspace:
How many 4-digit numbers can be formed, if repetition of digits is not allowed?
Answer: 120
Explanation :
Let us label the digits of 5-digit number as 4 3 2 1.
The rightmost digits i.e., 1st digit can be filled by either of 2 or 3 or 4 or 7 or 8, i.e., in 5 ways.
Now since repetition is not allowed, 2nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.
Similarly, 3rd digit can be filled by remaining 3 numbers in 3 ways.
and 4th digit can be filled in 2 days.
⇒ Total number of such 5-digit numbers = 5 × 4 × 3 × 2 = 120.
Hence, 120.
Workspace:
How many 4-digit even numbers can be formed, if repetition of digits is not allowed?
Answer: 72
Explanation :
Let us label the digits of 5-digit number as 5 4 3 2 1.
The rightmost digits i.e., 1st digit can be filled by either of 2 or 4 or 8, i.e., in 3 ways.
Now since repetition is not allowed, 2nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.
Similarly, 3rd digit can be filled by remaining 3 numbers in 3 ways.
and 4th digit can be filled in 2 days.
⇒ Total number of such 5-digit numbers = 3 × 4 × 3 × 2 = 72.
Hence, 72.
Workspace:
How many 4-digit numbers can be formed which are multiples of 4, if repetition of digits is not allowed?
Answer: 24
Explanation :
For a number to be divisible by 4, last 2-digits of the number should be divisible by 4.
Hence last two digits of the number should be either 32 or 72 or 84 or 48 i.e., 4 ways.
For each of these 4 ways, rest of the two digits can be filled by remaining 3 digits in 3 × 2 = 6 ways.
⇒ Total number of 4-digit numbers that can be formed which are multiples of 4 is 4 × 6 = 24.
Hence, 24.
Workspace:
How many 4-digit numbers can be formed which are multiples of 3, if repetition of digits is not allowed?
Answer: 24
Explanation :
For the number to be divisible by 3, the sum of the digits should be divisible by 3.
Here sum of 5 digits is 2 + 3 + 4 + 7 + 8 = 24.
We need to select 4 numbers (or leave out 1 number) such that sum is divisible by 3.
The only way sum would be divisible by 3 is when we select 2, 4, 7 and 8.
∴ We have to form a 4-digit number using 2, 4, 7 and 8 without repetition.
⇒ Number of ways = 4! = 24.
Hence, 24.
Workspace:
How many 4-digit numbers can be formed such that 2 always comes before 8, if repetition of digits is not allowed?
Answer: 60
Explanation :
Total number of 4-digit numbers without repetition = 5 × 4 × 3 × 2 = 120.
Out of these 120 numbers, half of them will have 2 before 8 and the other half will have 8 before 2.
∴ Number of 4-digit numbers that can be formed such that 2 always comes before 8 = 120/2 = 60
Hence, 60.
Workspace:
How many 5-digit numbers can be formed having exactly 1 digit between 3 & 7, if repetition of digits is not allowed?
Answer: 36
Explanation :
Let us label the 5 digits in the number i.e., 5 4 3 2 1.
We need to place 1 digit between 3 and 7.
Hence, we can place 3 and 7 at the following positions:
5 and 3, 4 and 2 or 2 and 1 i.e., 3 ways.
For each of these 3 places, 3 and 7 can be arranged in 2 ways, hence total number of ways of arranging 3 and 7 = 3 × 2 = 6 ways.
Now, for each of these 6 ways, rest 3 digits can be arranged in 3! = 6 ways.
Hence, total number of 5-digit numbers that can be formed having exactly 1 digit between 3 & 7 = 6 × 6 = 36 ways.
Hence, 36.
Workspace:
How many 5-digit numbers can be formed such that 4 & 2 always come together, if repetition of digits is not allowed?
Answer: 48
Explanation :
Let us keep 4 and 2 together and call it X.
Now we have the following digits 3, 7, 8 and X.
These 4 digits can be arranged in 4! = 24 ways.
For each of these 24 ways, there will be 2! Internal arrangement for 2 and 4 in X.
∴ Total number of 5-digit numbers that can be formed such that 4 & 2 always come together = 24 × 2 = 48.
Hence, 48.
Workspace:
What is the sum of all the 5-digit numbers that can be formed, if repetition of digits is not allowed?
Answer: 6399936
Explanation :
If 2 is at the unit's place, the remaining digits 3, 4, 7 and 8 can be arranged in 4! ways. Hence, there are exactly 4! 5-digit numbers using the given digits.
Similarly, there are exactly 4! 5-digit numbers using the given digits. Similarly for other digits.
Hence the total sum of digits at unit's place for all numbers possible is 4! × (2 + 3 + 4 + 7 + 8) × 1
Similarly, the total sum of digits at ten's place for all numbers possible is 4! × (2 + 3 + 4 + 7 + 8) ×
10
Similarly, the total sum at
hundred's place = 4! × (2 + 3 + 4 + 7 + 8) × 102
thousand's place = 4! × (2 + 3 + 4 + 7 + 8) × 103
ten thousand's place = 4! × (2 + 3 + 4 + 7 + 8) × 104
Hence, the required sum =
4! × (2 + 3 + 4 + 7 + 8)
+ 4! × (2 + 3 + 4 + 7 + 8) × 10
+ 4! × (2 + 3 + 4 + 7 + 8) × 102
+ 4! × (2 + 3 + 4 + 7 + 8) × 103
+ 4! × (2 + 3 + 4 + 7 + 8) × 104
= 4! × (2 + 3 + 4 + 7 + 8) × (11111)
= 24 × 24 × 11111
= 63,99,936.
Hence, 6399936.
Workspace:
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