If you have $9$ apples, you can choose $0, 1, 2, 3, ... 9$ of them. So there are $10$ ways to choose. Similarly for oranges, you have up to $7$ ways to choose. By the fundamental counting principle, you have $10\times7 = 70$ possible ways of choosing fruits (or no fruit). There is only one way to choose no fruit so we subtract $1$.
If you've used generating functions, you could also see it but it would also be a hassle to expand.
Also, have you ever encountered a problem such as "How many divisors does x have? (x non-prime)" It is similar to that where you add $+1$ to each power of the prime factorization when multiplying them.
If each apple and each orange were distinguishable, there would again be only one way to choosing nothing.
If the $9$ apples were distinguishable and the $6$ oranges were distinguishable, I believe this is the same as considering $9 + 6 = 15$ different objects.
How many ways are there to choose from $15$ different objects? There are $2^{15}$ such ways (including not choosing anything). Then $2^{15}-1$ is gives you the number of ways in which you have to choose at least one fruit. This is analogous to a power set.
You have to distribute each type of fruit separately.
Let $a_k$, $1 \leq k \leq 3$, be the number of apples distributed to the $k$th person. Then
$$a_1 + a_2 + a_3 = 5 \tag{1}$$
The number of ways the apples can be distributed is the number of solutions of equation 1 in the nonnegative integers. A particular solution corresponds to the placement of two addition signs in a row of five ones. For instance,
$$1 1 + 1 + 1 1$$
corresponds to the solution $a_1 = 2$, $a_2 = 1$, and $a_3 = 2$, while
$$+ 1 1 1 + 1 1$$
corresponds to the solution $a_1 = 0$, $a_2 = 3$, and $a_3 = 2$. Therefore, the number of such solutions is the number of ways two addition signs can be inserted into a row of five ones, which is
$$\binom{5 + 2}{2} = \binom{7}{2}$$
since we must choose which two of the seven symbols (five ones and two addition signs) will be addition signs.
The number of ways the mangoes can be distributed is the number of solutions in the nonnegative integers of the equation
$$m_1 + m_2 + m_3 = 4 \tag{2}$$
where $m_k$, $1 \leq k \leq 3$, is the number of mangoes distributed to the $k$th person. The number of such solutions is
$$\binom{4 + 2}{2} = \binom{6}{2}$$
The number of ways the oranges can be distributed is the number of solutions in the nonnegative integers of the equation
$$o_1 + o_2 + o_3 = 3 \tag{3}$$
where $o_k$, $1 \leq k \leq 3$, is the number of oranges distributed to the $k$th person. The number of such solutions is
$$\binom{3 + 2}{2} = \binom{5}{2}$$
To find the number of ways of distributing the three types of fruit, multiply the above results.
Best Answer
Your solution is correct. We can confirm it with another approach.
The number of subsets of a set with $n$ elements is $2^n$ since each element is either included in a subset or it is not.
Since there are $5$ apples, there are $2^5$ subsets of apples. Of these, $\binom{5}{0} + \binom{5}{1} = 1 + 5 = 6$ have fewer than two apples. Thus, $2^5 - 6 = 32 - 6 = 26$ have at least two apples.
Since there are four mangoes, there are $2^4$ subsets of mangoes. Of these, $\binom{4}{0} + \binom{4}{1} = 1 + 4 = 5$ have fewer than two mangoes. Thus, $2^4 - 5 = 16 - 5 = 11$ have at least two mangoes.
Since there are three bananas, there are $2^3$ subsets of bananas. Of these, $\binom{3}{0} + \binom{3}{1} = 1 + 3 = 4$ have fewer than two bananas. Thus, $2^3 - 4 = 8 - 4 = 4$ have at least two bananas.
Hence, the number of ways of selecting at least two fruits of each variety is $$26 \cdot 11 \cdot 4 = 1144$$ as you found.