Let me say what I think the rules are:
- At least one coin must be present.
- No more than one of each coin denomination must be present.
First we look at the three coins of lowest denomination: 1 paisa, 2 paisa, 3 paisa.
The sums of money from just these coins are: 1, 2, 1+2, 3, 3+1, 3+2, 3+2+1 making R0.01, R0.02, R0.03, R0.03, R0.04, R0.05, R0.06 respectively. There are 7 combinations, but only 6 discrete sums. 3 paisa can be made by combining 1 and 2 or by using the 3 paisa coin.
For the remaining 5 coins there are 32 combinations. Note that we cannot get the same sum from combining different coins for the highest denomination coins. There are no overlaps, unlike in the lowest denomination, where there is an overlap. Basically, what we are really doing is either using or not using the coin, so we have a binary system: the coin is OFF or ON, giving us 2⁵=32 combinations of OFF or ON, that is using none of the 5 coins up to using them all. So the total number of combinations is 32 times the number of combinations of the 3 other coins, and we know that the latter is 6 plus 1=7, because we must include not using any of the lowest denomination coins. In other words, we have R0.01-R0.06 (6 sums), then R0.10-R0.16 (7 sums). So we appear at first to have 32×7=224 different sums. We actually have 31×7+6=223.
Therefore we conclude that there are 223 different sums.