We can assume there are 103 things. Out of these 100 are coins and 3 are partitions (or sticks as mentioned). Now we see the number of ways to select the 3 things to appoint them as 'Partion'(or stick) which has 103C3 ways.
2. As per the restrictions given:
I assumed the partion number 1 is attached with 1 coin to its left and the partion number 2 is attached to two coin to its left. So, now considering these as one 'thing'. We now have 100 things to choose 3 things to appoint them as partion. Which has 100C3 ways.
My approach to this problem was like:
1. If there were no restrictions:
We can assume there are 103 things. Out of these 100 are coins and 3 are partitions (or sticks as mentioned). Now we see the number of ways to select the 3 things to appoint them as 'Partion'(or stick) which has 103C3 ways.
2. As per the restrictions given:
I assumed the partion number 1 is attached with 1 coin to its left and the partion number 2 is attached to two coin to its left. So, now considering these as one 'thing'. We now have 100 things to choose 3 things to appoint them as partion. Which has 100C3 ways.