Combinatorics: Distributing coins among three persons 
Seven different coins are to be divided among three persons. If no two of the persons receive the same number of coins but each receive at least one coin & none is left over, then the number of ways in which the division may be made is:
(A) 420
(B) 630
(C) 710
(D) None of these

The answer given is option (B)
This is how I solved it
Using inclusion-exclusion principle, distribution of 7 different coins into 3 different groups (of unknown size), such that no group is empty.
$$3^7-^3C_1(3-1)^7+^3C_2(3-2)^7=1806$$ But this also includes the counting of groups with any two same number of coins, thus we need to subtract that from the above term, let it be $K$.
The only possible case with "If no two of the persons receive the same number of coins" is 2 same, 2 same, 1 different, these are $$(3,3,1), (2,2,3),(1,1,5)$$ Thus the corresponding arrangements are:-$$\frac{7!}{3!3!1!} , \frac{7!}{2!2!3!}, \frac{7!}{1!1!5!}$$Sum of these three $K=392$
Thus answer $=1806-392=1414$
But the answer given is option (B). Where am I wrong? How would you solve the problem?
 A: I would reason as follows :
The only distribution that satisfies the conditions is $(1, 2, 4)$.
This gives rise to $\binom{7}{4} \times \binom{3}{2} \times \binom{1}{1}$ arrangements : first we choose the 4-coin group, then among the 3 remaining coins we choose the 2-coin group, then among the 1 coin left we "choose" the 1-coin group. That gives 105 possibilities.
Now that we have separated the coins into three unequal (in terms of cardinal), non-empty groups, we assign these three groups to the three people : there are $3 !$ possibilities.
Total is therefore $105 \times 6 = 630$.
A: The only possible split up for $7$ as a sum of $3$ distinct positive integers with summation $7$ is: $$7=1+2+4$$
There are $3!$ permutations of $(1,2,4)$.
And if e.g. the first person gets $1$, the second gets $2$ and the third gets $4$ coins then there are $\frac{7!}{1!2!4!}$ distinct ways to divide the coins.
So eventually we find:$$3!\times\frac{7!}{1!2!4!}=630$$ possibilities.
A: I got the error in my method so I thought to write answer myself.
As I already mentioned the three cases that needs to be subtracted and named it $K$, the problem was that I already selected that first two groups will be same and the third will be different, but in reality we need to arrange these as $\frac{3!}{2!1!}=3$, hence the terms 
$$\frac{7!}{3!3!1!} , \frac{7!}{2!2!3!}, \frac{7!}{1!1!5!}$$needs to be multiplied by three and then added thus $K$ will be $3\times392=1176$
And hence the answer will be $1806-1176=630$
A: The conditions that everyone receives a different number, but no-one receives none, mean that the only possibility is that the coins are split $1,2,4$ in some order among the people.
There are $7$ ways to choose the single coin, and $\binom 62$ ways to choose a pair of coins from what's left, then $3!$ ways to assign the three people to the three sets of coins. So the answer is $7\times 15\times 6$.
