Question

31. Five digit number can be formed from 3, 1,7,0,9,5 divisible by 5 are :
c) 96
a) 120
d) 384
b) 216​

Answer 1

Answer:

Correct option is

B

216

The numbers are divisible by 5 is they end with either 0 or 5

Case 1:

Let the number ends with 0 i.e. unit's digit is 0

Now there are 5 choices for ten's place

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 0 is 5×4×3×2=120

Case 2:

Let the number ends with 5 i.e. unit's digit is 5

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 5 is 5×4×3×2=120

But the first digit can't be 0. So, we need to subtract that numbers which contains 0 as first digit.

If 0 is the first digit, then number of places left to be filled is 4 and that can be done in 4! ways.

So, the 5 digits numbers that can be formed which ends with 5 is 120−4!=96

Thus, the total number of five digit numbers divisible by 5 is 120+96=216

Answer 2

Answer:

The numbers are divisible by 5 is they end with either 0 or 5

Case 1:

Let the number ends with 0 i.e. unit's digit is 0

Now there are 5 choices for ten's place

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 0 is 5×4×3×2=120

Case 2:

Let the number ends with 5 i.e. unit's digit is 5

As the repetition is not allowed, 4 choices for hundred' place.

3 choices for thousand's place and 2 choices for the first digit.

So, the 5 digits numbers that can be formed which ends with 5 is 5×4×3×2=120

But the first digit can't be 0. So, we need to subtract that numbers which contains 0 as first digit.

If 0 is the first digit, then number of places left to be filled is 4 and that can be done in 4! ways.

So, the 5 digits numbers that can be formed which ends with 5 is 120−4!=96

Thus, the total number of five digit numbers divisible by 5 is 120+96=216