Answer should be $2^8 - 32 = 224$

Of Course we have to subtract some repeated/duplicated combinations from $2^8$(Which is the maximum possible combinations)

Now we will get duplicates by Only $1,2,3 \,\,Paisa\,\,Coins$

i.e. We need to find the Combinations where $3$ can be replaced with $1,2$

**APPROACH 1 : **

**The Idea is that We can always replace $1\,\,and\,\,2\,\,paisa \,\,Coins$ with the $3\,\,Paisa\,\,Coin$ and get the same Sum. So, When We replace $1\,\,and\,\,2\,\,paisa \,\,Coins$ with the $3\,\,Paisa\,\,Coin$, We are getting same sum But with One less Coin. hence we fall from $n \,\,coins $ in hand to $n-1$ coins in hand but with the same Sum. ** Bases on this Idea, The following answer is Proposed :

Let's find out such combinations :

**A.** Selecting Zero Coin //No problem so far

**B.** Selecting One Coin . //No problem here also

**C.** Selecting Two Coins : Coming to Selecting Two coins, We can see that when we select $2,1$ we get Sum as $3$ which is already there in $A$ Row.

**D**. Selecting Three Coins : When Two of these Three coins are $1,2\,\,Paisa\,Coins$ we can replace them with $3\,Paisa\,\,Coin$ and the resulting sum is already there in Row $C$ .... So, When Two of the Selected Three coins are $1,2$ we have $5$ possibilities to choose the third coin (We can not select $3 Paisa$ as the Third Coin)

**E.** Selecting Four Coins : When Two of these Four coins are $1,2 \,Paisa\,\,Coins$ we can replace them with $3\,Paisa \,\,Coin$ and the resulting sum is already there in Row $D$ .... So, When Two of the Selected Four coins are $1,2\,Paisa\,Coins$ we have $\binom{5}{2}$ possibilities to choose the Other Two coin (We can not select $3 Paisa$ as any of the rest Two Coin)

**F.** Selecting Five Coins : When Two of these Five coins are $1,2 \,Paisa\,\,Coins$ we can replace them with $3\,Paisa \,\,Coin$ and the resulting sum is already there in Row $E$ .... So, When Two of the Selected Five coins are $1,2\,Paisa\,Coins$ we have $\binom{5}{3}$ possibilities to choose the Other Three coins (We can not select $3 Paisa$ as any of the rest Three Coins)

**G.** Selecting Six Coins : When Two of these Six coins are $1,2 \,Paisa\,\,Coins$ we can replace them with $3\,Paisa \,\,Coin$ and the resulting sum is already there in Row $F$ .... So, When Two of the Selected Six coins are $1,2\,Paisa\,Coins$ we have $\binom{5}{4}$ possibilities to choose the Other Four coins (We can not select $3 Paisa$ as any of the rest Four Coins)

**H.** Selecting Seven Coins : When Two of these Seven coins are $1,2 \,Paisa\,\,Coins$ we can replace them with $3\,Paisa \,\,Coin$ and the resulting sum is already there in Row $G$ .... So, When Two of the Selected Seven coins are $1,2\,Paisa\,Coins$ we have $\binom{5}{5}$ possibilities to choose the Other Five coins (We can not select $3 Paisa$ as any of the rest Five Coins)

So, Number of Duplicate Cases are (From the above $A \,\,to\,\,H$ Points) = $1+5+10+10+5+1 = 32$

Hence We have Our answer = $256-32 = 224$

**APPROACH 2 : **

As @Soumya Suggested,

Repeated sums can be obtained only when either 3 paisa coin or (2+1)paisa coins is there in the selected coins.

$2^8$ sums can be possible and among them 2^5 (Cases where 3 paisa coin is not selected and (2+1) paisa both coins are selected) are repeated.

$2^8 - 2^5 = 224$

Thanks to @Soumya for Suggesting Such a Straight and Short way.