CRE 4 - Forming numbers using digits

Let us label the digits of 5-digit number as 5 4 3 2 1.

The rightmost digits i.e., 1^st 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, 2^nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.

Similarly, 3^rd digit can be filled by remaining 3 numbers in 3 ways.

∴ 4^th digit can be filled in 2 days.

and 5^th digit can be filled in 1 way.

⇒ Total number of such 5-digit numbers = 5 × 4 × 3 × 2 × 1 = 120.

Hence, 120.

Let us label the digits of 5-digit number as 5 4 3 2 1.

The rightmost digits i.e., 1^st 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, 2^nd digit can also be filled by any of these 5 numbers, i.e., in 5 ways.

Similarly, 3^rd, 4^th and 5^th 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.

Let us label the digits of 5-digit number as 4 3 2 1.

The rightmost digits i.e., 1^st 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, 2^nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.

Similarly, 3^rd digit can be filled by remaining 3 numbers in 3 ways.

and 4^th digit can be filled in 2 days.

⇒ Total number of such 5-digit numbers = 5 × 4 × 3 × 2 = 120.

Hence, 120.

Let us label the digits of 5-digit number as 5 4 3 2 1.

The rightmost digits i.e., 1^st digit can be filled by either of 2 or 4 or 8, i.e., in 3 ways.

Now since repetition is not allowed, 2^nd digit can be filled by any of the 4 remaining numbers, i.e., in 4 ways.

Similarly, 3^rd digit can be filled by remaining 3 numbers in 3 ways.

and 4^th digit can be filled in 2 days.

⇒ Total number of such 5-digit numbers = 3 × 4 × 3 × 2 = 72.

Hence, 72.

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 24 or 84 or 28 or 48 i.e., 6 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 6 × 6 = 36.

Hence, 36.

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.

2 and 8 should definitely be included, hence we need to select two more digits out of 3, 4 and 7.
This can be done in ³C₂ = 3 ways.
Number of ways of arranging any selection of 4 digits = 4! = 24

∴ Total number of 4-digit numbers without repetition = 3 × 4! = 72.

Out of these 72 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 = 72/2 = 36.

Hence, 36.

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.

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.

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.