Question

how many numbers can be formed from 1 2 3 4 5 (without repetition) when the digit at the unit’s place must be greater than that in the ten’s place?

Answer 1

total method without any restriction = 5*4*3*2*1=120
now for than tens place , only one method out of the set of two will be correct ex. 12 and 21 only one will be valid so answer will be 120/2=60

Answer 2

Answer:

The total number of possible ways are:   6+12+18+24 = 60

Step-by-step explanation:

There are total 5 digits to be used 1 2 3 4 5 (without repetition).

Its given that digit at the unit’s place must be greater than that in the ten’s place, so unit place cant be 1 as it is smallest from given set.

So if unit’s place is 2, the ten’s place can be 1 and we are left with 3 more places to be filled by remaining 3 digits, which can be done in 6 ways (3*2)

If unit’s place is 3, the ten’s place can be 1 or 2.  So, we are left with 3 more places to be filled by remaining 3 numbers, which can be done in 6 ways (3*2). So the total number of ways is 6*2 = 12 (12 ways for each of tens place being 1 or 2) .

If unit’s place is 4, the ten’s place can be 1 or 2 or 3.  So, we are left with 3 more places to be filled by remaining 3 numbers, which can be done in 6 ways (3*2). So the total number of ways is 6*3 = 18 (6 ways for each of tens place being 1 or 2 or 3)

If unit’s place is 5, the ten’s place can be 1 or 2 or 3 or 4.  So, we are left with 3 more places to be filled by remaining 3 numbers, which can be done in 12 ways (3*2). So the total number of ways is 6*4 = 24 (6 ways for each of tens place being 1 or 2 or 3 or 4)

so total number of possible ways are:  6+12+18+24 = 60