CRE 4 - Forming numbers using digits | Modern Math - Permutation & Combination

Answer: 120

Explanation :

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.

Answer: 3125

Explanation :

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.

Answer: 120

Explanation :

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.

Answer: 72

Explanation :

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.

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.

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.

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.

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.

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.

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.