Question

How many 3-digit
numbers can be
formed from the
digits 2, 3, 5, 8 and
9, without repetition,
which are exactly
divisible by 4?​

Answer 1

Step-by-step explanation:

Given that:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 8 and 9, without repetition, which are exactly divisible by 4?

To find:No of ways to arrange so that numbers are exactly divisible by 4

Solution:

From the given numbers 2,3,5,8 and 9

2 and 8 can occupy last (unit digits)

Reason: A number is divisible by 4 only if it's last two digits are divisible by 4.

Case 1: Fix 2 at unit place.

Now we have four numbers out of which we have to choose 2 digits for tenth and hundredth place

n= 4(3,5,8,9)

r= 2(only 2 place vacant)

No. of methods:

$^{4} P_2 = \frac{4!}{(4 - 2)!} \\ \\ = \frac{4!}{2!} \\ \\ = \frac{4\:\times\:3\times\:2!}{2!} \\ \\=12\:methods$

Out of which the numbers which have 8 at tenth place are not divisible by 4

(382,582,982)= 3 numbers should remove from 12 ways

Total ways to arrange = 12-3= 9 ways

Case2: Fix 8 at unit place.

Now we have four numbers out of which we have to choose 2 for tenth place

because 38,58,98 are not divisible by 4

n= 3(3,5,9)

r= 1(only 1 place vacant)

No. of methods:

$^{3} P_1 = \frac{3!}{(3 - 1)!} \\ \\ = \frac{3!}{2!} \\ \\ = 3\:ways$

Total possible ways:

No. of ways from case1+ No. of ways from case2

=9+3

= 12

Total 3 digits numbers formed from given 5 digits are 12 numbers, which are divisible by 4.

Hope it helps you.