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?
Best Answer
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 $105 \times 6 = 630$.