The problem you're asking about deals with combinations of objects from different categories. Specifically, you want to know how many ways there are to choose a certain number of objects $n_1, n_2, \dots, n_k$ from $k$ types of objects, where the total number of objects of each kind is $N_1, N_2, \dots, N_k$, and you're selecting $n_i$ objects of the $i$-th kind.

Case 1: $n_i \leq N_i$ for all $i$

If you are choosing $n_i$ objects from $N_i$ available objects of the $i$-th kind (and $n_i \leq N_i$ for all $i$), the number of ways to choose these objects is simply the product of combinations for each kind of object:

$\text{Total number of ways} = \binom{N_1}{n_1} \times \binom{N_2}{n_2} \times \cdots \times \binom{N_k}{n_k}$

Where $\binom{N_i}{n_i}$ represents the binomial coefficient, or the number of ways to choose $n_i$ objects from $N_i$ objects of the $i$-th kind, which is given by:

$\binom{N_i}{n_i} = \frac{N_i!}{n_i!(N_i - n_i)!}$

Thus, the total number of ways to choose the objects from all categories is:

$\prod_{i=1}^{k} \binom{N_i}{n_i}$

Case 2: $n_i > N_i$ for some $i$

If $n_i > N_i$ for any $i$, the problem becomes invalid because you cannot select more objects than are available in any category. In this case, there is no valid way to choose the objects, so the total number of ways is zero. Therefore:

$\text{Total number of ways} = 0 \quad \text{if} \quad n_i > N_i \quad \text{for any} \, i.$

Summary:

1. If $n_i \leq N_i$ for all $i$, the number of ways to choose the objects is: $\prod_{i=1}^{k} \binom{N_i}{n_i}$

2. If $n_i > N_i$ for any $i$, the number of ways is: $0$

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Would you like more details or further clarifications on this?

Here are 5 related questions that build on this information:

1. How do you calculate the binomial coefficient $\binom{N_i}{n_i}$?
2. What is the difference between combinations and permutations?
3. How does this change if the order of objects matters (i.e., we are dealing with permutations)?
4. How does the total number of ways change if some objects are indistinguishable?
5. Can we apply this formula to infinite sets? If so, how?

Tip: For large values of $N_i$, you can use approximations like Stirling's formula to simplify factorial calculations in binomial coefficients.