Question

How many 5 digit numbers can be formed using the digits 0,1,2,3,4,5 which are divisible by 5 without repetion ?

Answer 1

Answer:A no. is divisible by 5 only when it’s unit’s digit is either 0 or 5.

So, Let’s break our question in 2 parts

Step-by-step explanation:

PART-1: When No. ends with “0”.

_ _ _ _ 0. (Where ‘_’ denotes blank spaces in which no. is to be filled)

Basically we have fixed the unit’s Digit as “0”. So now we are left with 5 digits

Now the First Blank space have 5 Options of Digits which it can acquire.

The Second Place has 4 Options of Digits. (As Digits can’t be repeated).

The Third Place has 3 Options of Digits.

The Fourth Place has 2 Options of Digits.

So total no. possible will be:

5*4*3*2=5!

5!=120.

PART-2: When No. ends with “5”.

_ _ _ _ 5.

Now the first Blank space can’t acquire “0” as it won’t be a 5-Digit no. then.

So the First Blank space have 4 choices.(i.e. 1,2,3 or 4.)

The Second Place also has 4 choices. (i.e. 0 and the other three no. left after filling first blank.)

The Third Place has 3 choices.

And The Fourth Place has any 2 choices.

so overall in this part no. of possible Numbers that can be formed will be

4*4*3*2=96.

Now Add the possible no. of ways of Part-1 and Part-2.

so, The Total No. of Possible Digits will be:

120+96=216.

Ans: 216.

Note: This is the most basic method of finding possible no. of Numbers Forming under certain given Conditions. I actually Don’t know the Method of Solving this question using PnC. Hope you understood What I did in the question.)

THANK YOU.