Permuting fruits in a row I have the following homework:

Assume we have $7$ apples, $4$ oranges, and $1$ banana. The apples and oranges are identical separately. Please count the number of all permutations for each of the following cases:  (a) put these fruits in a row;  (b) put these fruits in a circle; (orientation matters, i.e., clockwise and counter-clock are different.)  (c) put these fruits in a row where the banana cannot be adjacent to any oranges.


For part a, I think the answer is
$$\frac{12!}{7!4!}=3960$$
For part b, I think the answer is
$$\frac{(12-1)!}{7!4!}=330$$
However, for part c, I don't even know how to approach this. I started by thinking about the situation the banana need to be adjacent to the oranges, after that, I will subtract it from the answer in part a which is the possible permutations. However, I don't know how to do that. Could anyone please help?
 A: Your first two answers are correct.
For the third question, your strategy is sound.
Suppose a banana has an orange to its immediate left.  Then the orange-banana pair must begin in one of the first $11$ positions.  That leaves seven apples and three oranges to distribute to the remaining $10$ positions, which can be done in $\binom{10}{3}$ ways.  Hence, there are
$$11\binom{10}{3}$$
arrangements in which a banana has an orange to its immediate left.
By symmetry, there are
$$11\binom{10}{3}$$
arrangements in which a banana has an orange to its immediate right.
However, if we subtract these amounts from the total, we will have subtracted each arrangement in which a banana is adjacent to two oranges twice, once when we subtracted arrangements in which there was a banana to its immediate left and once when we subtracted arrangements in which there was a banana to its immediate right.  We only want to subtract such arrangements once, so we must add them to the total.
If a banana is adjacent to two oranges, then the orange-banana-orange block must begin in one of the first $10$ positions.  That leaves us with seven apples and two oranges to distribute to the remaining nine positions, which can be done in $\binom{9}{2}$ ways.  Hence, there are
$$10\binom{9}{2}$$
such arrangements.
By the Inclusion-Exclusion Principle, the number of arrangements of seven indistinguishable apples, four indistinguishable oranges, and a banana in a row in which the banana is not adjacent to an orange is
$$\binom{12}{7, 4, 1} - 2 \cdot 11\binom{10}{3} + 10\binom{9}{2}$$
where the multinomial coefficient
$$\binom{12}{7, 4, 1} = \frac{12!}{7!4!1!}$$
A: B: banana,A: apple,
O: orange
For (c), see that, we have 12 positions to fill in total where banana and orange cannot be adjacent to each other.
Since there is just 1 banana, you get 2 cases, it'll either be at the end or somewhere in the middle. So,
Case 1: Banana lies at either of the two ends of the row
Positioning will be something like,
B A _ _ _ _ _ _ _ _ _ _
or
_ _ _ _ _ _ _ _ _ _ A B
Here, the banana and an apple's position will be fixed, and, rest of the 6 apples and 4 oranges can be adjusted in the 10 places in any possible way.
So number of ways for such an arrangement becomes 2[1.1.10!/6!4!]
Case 2: Banana lies somewhere in between.
Positioning will be like, ....A B A....
Here, 2 apples and the banana will have fixed positions so you can consider them as a single unit say X=ABA.
eg., _ _ X _ _ _ _ _ _ _
i.e.  _ _ A B A _ _ _ _ _ _ _
Note that X can be placed at any of the 10 places in the arrangement.
So, now you need to fill up the remaining 9 positions with 5 apples and 4 oranges.
number of ways to do that= 10[1.9!/5!4!]
Therefore,
total number of cases for (c) are [(2.10!/6!4!) + (10!/5!4!)] = 1680
