Question

How many different numbers of six digits (without repetition of digits) can be formed from the digits 3,1,7,0,9,5
I.How many of them will have 0 in the unit’s place?
II.How many of them are divisible by 5?
III.How many of them are not divisible by 5?

Answer 1

Hint: We are given a problem in which there are 6 different digits given such as to form a six digit number. Then in the second sub question that number should be divisible by 5 that is on the last two places can be filled by either 0 or 5. And in the third sub question we need to find the numbers that are not divisible by 5. That is, the last two places should have any digit other than 0 and 5.

Complete step by step solution:
We will solve the problem step by step.
I)First let's find the numbers that can be formed.
Now we cannot place 0 on the highest place value because that will not be the six digit number. So we have five other places for it.
Now one of the remaining 5 numbers can be placed in 5 ways. Similarly one by one the numbers will be placed. So the total numbers will be,
\[5 \times 5!\]
That is \[5 \times 120 = 600\]
600 different numbers will be there.
II) They may have 5 in the last place and as above we will have 5! = 120 ways. These will also include numbers which will have zero in the first place. Therefore the numbers having zero in 1st and 5 in last place will be 4!
Therefore 6 digit numbers having 5 in the end will be
$5! -4! = 120 - 2.4 = 96$.
Therefore the total number of 6 digit numbers divisible by 5 is
120 + 96 = 216.
Only 216 numbers are there; those are divisible by 5.
III) At last the numbers that are not divisible by 5 can be found directly by subtracting the numbers that are divisible from total numbers.
So that is \[600 - 216 = 384\]
Thus we completed the solution.

Note: Note that placing zero on the highest place value will not fulfill the condition of the 6 digit number mentioned. And in order to find the number not divisible by 5. we will not shuffle again. we will just remove the numbers that are divisible.
The other divisible numbers can also be found just by using the divisibility tests of the respective numbers requirement.