I can analyze the math problems from your image. Here are the transcribed problems along with their solutions:

1. If there are 12 teams in a basketball tournament and each team must play every other team, how many elimination games will there be?

This is a combination problem where we choose 2 teams out of 12 to form a pair: $\binom{12}{2} = \frac{12 \times 11}{2} = 66 \text{ games}.$

2. If there are 7 distinct points on a plane with no three of which are collinear, how many different polygons can be formed?

To form a polygon, we must select at least 3 points: $\binom{7}{3} + \binom{7}{4} + \binom{7}{5} + \binom{7}{6} + \binom{7}{7}.$ Calculating each term: $\binom{7}{3} = 35, \, \binom{7}{4} = 35, \, \binom{7}{5} = 21, \, \binom{7}{6} = 7, \, \binom{7}{7} = 1.$ Total: $35 + 35 + 21 + 7 + 1 = 99 \text{ polygons}.$

3. How many different sets of 5 cards each can be formed from a standard deck of 52 cards?

This is another combination problem where we select 5 cards from 52: $\binom{52}{5} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2598960 \text{ sets}.$

4. In a 10-item mathematics problem-solving test, how many ways can you select 5 problems to solve?

We select 5 problems out of 10: $\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \text{ ways}.$

5. In problem number 4, how many ways can you select the 5 questions if you are required to answer question number 10?

If question 10 must be included, we only need to select 4 more questions from the remaining 9 questions: $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \text{ ways}.$

6. In how many ways can a committee of 5 be formed from 5 juniors and 7 seniors if the committee must have 3 seniors?

We need to select 3 seniors from 7 and 2 juniors from 5: $\binom{7}{3} \times \binom{5}{2}.$ Calculating: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35, \quad \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10.$ Total: $35 \times 10 = 350 \text{ ways}.$

7. From a population of 50 households, in how many ways can a researcher select a sample with a size of 10?

We select 10 households from 50: $\binom{50}{10} = \frac{50 \times 49 \times 48 \times \dots \times 41}{10 \times 9 \times \dots \times 1}.$ This value is very large, approximately: $10272278170 \text{ ways}.$

8. A box contains 5 red balls, 7 green balls, and 6 yellow balls. In how many ways can 6 balls be chosen if there should be 2 balls of each color?

We choose 2 red balls from 5, 2 green balls from 7, and 2 yellow balls from 6: $\binom{5}{2} \times \binom{7}{2} \times \binom{6}{2}.$ Calculating: $\binom{5}{2} = 10, \, \binom{7}{2} = 21, \, \binom{6}{2} = 15.$ Total: $10 \times 21 \times 15 = 3150 \text{ ways}.$

Would you like a detailed explanation for any of these problems?

Related Questions:

1. How does the formula for combinations $\binom{n}{r}$ work conceptually?
2. Can you explain the difference between permutations and combinations?
3. What is the total number of subsets that can be formed from 7 points?
4. How can we calculate large combinations like $\binom{50}{10}$ more efficiently?
5. What changes in the answer if there were 8 seniors and 6 juniors in problem 6?

Tip:

Always check if restrictions (like a required member or condition) simplify the problem before calculating combinations.