To solve this problem, we need to determine the number of ways to choose 10 articles from 10 bags, where each bag contains different numbers of articles. Specifically, bag \( B_1 \) contains 21 articles, \( B_2 \) contains 22, and so on, up to bag \( B_{10} \), which contains 30 articles.
We can approach this problem using combinations. The task requires selecting a total of 10 articles from the 10 bags, but we need to consider two main constraints:
- Selecting articles such that at least one article is picked from each bag.
- Ensuring a total of 10 articles is achieved.
First, we find the number of ways to choose 1 article from each of the 10 bags, which ensures at least one article is selected from each. Each choice is independent of the others, and this initial step involves no choice actually because each bag must contribute at least one article.
After removing one article from each bag (totaling 10 articles), we are left with choosing 0 more articles to complete the requirement of 10 articles in total. So the number of ways to select additional 0 articles from the remaining ones is given by:
The problem then boils down to distributing these additional articles among the bags with their remaining capacity. However, here we aim to simplify the computational burden by using the direct solution approach based on logical deduction:
Each bag can contribute more than one or the total number of articles in it minus one that was compulsorily taken, which is solved computationally as follows:
The total sum of remaining articles in all bags after taking one article from each:
The expression to compute available selections for extra 0 (which is a mathematical style adapted here) or contribute of surplus is:
\(^{30}C_{20} + ^{20}C_{10}\)
This calculation is from total remaining capacity to be distributed fulfilling combination criteria. Now let us understand the options:
- \(^{31}C_{20} + ^{21}C_{10}\) and other incorrect ones are not logically splitting per the restriction of articles being chosen per bag attributed prior hence does not compute into probability space correctly.
- Our established calculation more explicitly aligns with the right constraints ensuring minimum article selection from each, accurately fulfilling the restrictions.
Thus, the correct option is: \(^{30}C_{20} + ^{20}C_{10}\).
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