How many numbers can be formed?

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

Hint: how many numbers can be formed at all, and what proportion of those have the digit in the units place bigger than that in the tens place?

Answer 2

we take 1 as in the tenth place digit then in units place 2,3,4 and 5 can only be used. so the first 3 digits from left can be parmuted using any 3 digits. So number of possible ways are 3! total ways=4*3! for 2,3*3! for 3,2*3! for4,1*3! therefore total numbers will be=(4+3+2+1)*3!=60

Answer 3

total methods without any restriction $= 5\cdot4\cdot3\cdot2\cdot1 = 120$

Now for units place to be greater than tens place , only one method out of the set of two will be correct.

Ex$:$ $12$ & $21$ only one will be valid so answer will be $\dfrac{120}{2} = 60$

So, $60$ numbers can be formed using all of $1,2,3,4,5($without repetition$),$when the digit at the units place must be greater than that in the tenth place.