Question number 7.




Suppose that 40 batteries are shipped to an auto parts store, and that 3 of those are defective.  A fleet manager then buys 7 of the batteries from the store.  In how many ways can at least 2 defective batteries be included in the purchase?


 2,065,414




 1,307,691




 1,373,736




 1,974,024




 1,241,646




 None of the above.



Question number 8.




A classroom of children has 18 boys and 20 girls in which five students are chosen to do presentations.  In how many ways can the five students be chosen so that more boys than girls are selected?


 163,608




 216,240




 61,200




 224,808




 8,568




 None of the above.



Question number 9.




A firm has 25 senior and 26 junior partners.  A committee of three partners is selected at random to represent the firm at a conference.  In how many ways can at least one of the junior partners be chosen to be on the committee?


 237,250




 18,525




 2,295,930




 26




 975




 None of the above.



### Question 7

You need to find how many ways at least **two defective batteries** can be included in the purchase of 7 batteries from a total of 40, where 3 are defective.

This is a problem of combinations, with a focus on "at least two defective" batteries. 

1. **Total number of ways to choose 7 batteries** (without any restriction):
   \[
   \text{Total combinations} = \binom{40}{7}
   \]

2. **Number of ways with no defective batteries** (choosing all 7 from the 37 non-defective ones):
   \[
   \text{No defective} = \binom{37}{7}
   \]

3. **Number of ways with exactly 1 defective battery** (choose 1 defective from 3, and 6 non-defective from 37):
   \[
   \text{1 defective} = \binom{3}{1} \times \binom{37}{6}
   \]

4. **The number of ways with at least 2 defective batteries** will be:
   \[
   \text{At least 2 defective} = \binom{40}{7} - \left( \binom{37}{7} + \binom{3}{1} \times \binom{37}{6} \right)
   \]

### Question 8

You need to select 5 students out of 18 boys and 20 girls, where **more boys than girls** are selected. The possible selections are:

- 3 boys, 2 girls
- 4 boys, 1 girl
- 5 boys, 0 girls

For each case, calculate the combinations:

1. **3 boys, 2 girls**:
   \[
   \binom{18}{3} \times \binom{20}{2}
   \]

2. **4 boys, 1 girl**:
   \[
   \binom{18}{4} \times \binom{20}{1}
   \]

3. **5 boys, 0 girls**:
   \[
   \binom{18}{5} \times \binom{20}{0}
   \]

Add all of these up to get the total number of ways to select more boys than girls.

### Question 9

The problem asks for the number of ways to select **at least one junior partner** in a committee of 3 from a group of 25 senior and 26 junior partners.

1. **Total number of ways to form the committee without restrictions**:
   \[
   \text{Total combinations} = \binom{51}{3}
   \]

2. **Number of ways to select only senior partners** (choosing all 3 from 25):
   \[
   \text{Only senior} = \binom{25}{3}
   \]

3. **Number of ways to have at least one junior partner**:
   \[
   \text{At least 1 junior} = \binom{51}{3} - \binom{25}{3}
   \]

---

Would you like detailed steps for any of the questions or have any other queries?

Here are 5 relative questions to practice:

1. How do you compute combinations involving larger groups?
2. What is the general formula for solving "at least" type problems?
3. How does inclusion-exclusion principle help in counting?
4. How would you approach probability with similar counting principles?
5. How do factorials relate to permutations and combinations?

**Tip**: For large combination calculations, simplify factorials by canceling out terms to reduce computational complexity.

Suppose that 40 batteries are shipped to an auto parts store, and that 3 of those are defective. A fleet manager then buys 7 of the batteries from the store. In how many ways can at least 2 defective batteries be included in the purchase?

Question number 8.

A classroom of children has 18 boys and 20 girls in which five students are chosen to do presentations. In how many ways can the five students be chosen so that more boys than girls are selected?

Question number 9.

A firm has 25 senior and 26 junior partners. A committee of three partners is selected at random to represent the firm at a conference. In how many ways can at least one of the junior partners be chosen to be on the committee?

Solve combinatorics problems involving selecting defective batteries, students for presentations, and partners for a committee. Learn how to calculate combinations using formulas like nCr and the Inclusion-Exclusion Principle. Suitable for students in Grades 10-12.

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