Permutation and Combination

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute =

Every number between 1000 and 2000, which is divisible by five and which can be formed by the given digits, must contains 5 in unit's place and 1 in thousand's place. Thus we are left with four digits out of which we are to place two between 1 and 5, which can be done in 4P2 = 12 ways. Hence, 12 numbers can be formed.

Rahul can occupy any one of the two corner positions and therefore, Rahul can occupy a position in two ways.

The other four people can seat themselves in 4! ways, that is, 24 ways.

Total ways = 2 * 24 = 48 ways.

1. i. Since repetition of number is allowed and there are five numbers which can be used to form the necessary 3-digit numbers we can have five numbers that can fill the ones, tens and hundreds place.

So, total number of possible 3-digit number = 5 * 5 * 5 = 125

ii. Since repetition of number is not allowed. There are total 5 numbers which can fill the ones places then 4 and 3 numbers which can fill the tens and hundreds place simultaneously.

So, total number of possible 3-digit number = 3 * 4 * 5 = 60