Algebra 4 Permutations & Combinations Worksheet Name: 1 ... - Free Printable

Here are the step-by-step solutions for each problem on the worksheet.

1. Fifteen students ask to visit... In how many ways can ten time slots be assigned?
* Logic: We have 15 distinct students and 10 distinct time slots. Since the time slots are different (e.g., 9:00 AM is different from 10:00 AM), the order matters. This is a permutation problem where we choose 10 people out of 15 and arrange them.
* Calculation: $P(15, 10) = \frac{15!}{(15-10)!} = 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6$.
* Result: $10,897,286,400$

2. The prom committee has four sites available for the banquet and three sites for the dance. How many arrangements are possible?
* Logic: You need to choose one site for the banquet AND one site for the dance. These are independent choices, so we multiply the number of options for each.
* Calculation: $4 \text{ (banquet sites)} \times 3 \text{ (dance sites)} = 12$.
* Result: $12$

3. How many different teams of 11 players can be chosen from a soccer squad of 16?
* Logic: The order in which you pick the players does not matter; being on the team is what counts. This is a combination problem.
* Calculation: $C(16, 11) = \frac{16!}{11!(16-11)!} = \frac{16!}{11!5!}$.
* This simplifies to $\frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1}$.
* $5 \times 3 = 15$, so cancel 15.
* $4 \times 2 = 8$, and $16 / 8 = 2$.
* Remaining: $2 \times 14 \times 13 \times 12 = 4,368$.
* Result: $4,368$

4. Suppose you find seven articles... In how many ways can you choose five articles to read?
* Logic: Order doesn't matter here (reading article A then B is the same set as reading B then A). This is a combination.
* Calculation: $C(7, 5) = \frac{7!}{5!(7-5)!} = \frac{7!}{5!2!}$.
* Simplify: $\frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21$.
* Result: $21$

5. For a band camp, you can choose two or three roommates from a group of 25 friends. In how many ways can you choose?
* Logic: You can either choose 2 friends OR choose 3 friends. We calculate both possibilities and add them together.
* Calculation:
* Choose 2: $C(25, 2) = \frac{25 \times 24}{2} = 300$.
* Choose 3: $C(25, 3) = \frac{25 \times 24 \times 23}{3 \times 2 \times 1} = 25 \times 4 \times 23 = 2,300$.
* Total: $300 + 2,300 = 2,600$.
* Result: $2,600$

6. A salad bar offers eight choices of toppings... choose four or five toppings?
* Logic: Similar to #5, we calculate the ways to choose 4 and the ways to choose 5, then add them. Order doesn't matter.
* Calculation:
* Choose 4: $C(8, 4) = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} = 70$.
* Choose 5: $C(8, 5) = C(8, 3)$ (since choosing 5 to keep is same as choosing 3 to leave out).
* $C(8, 3) = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$.
* Total: $70 + 56 = 126$.
* Result: $126$

7. In how many ways can four distinct positions for a relay race be assigned from a team of nine runners?
* Logic: The positions are "distinct" (e.g., 1st leg, 2nd leg). Therefore, who runs which leg matters. This is a permutation.
* Calculation: $P(9, 4) = 9 \times 8 \times 7 \times 6$.
* $9 \times 8 = 72$.
* $7 \times 6 = 42$.
* $72 \times 42 = 3,024$.
* Result: $3,024$

8. A committee must choose 3 finalists from 15 scholarship candidates.
* Logic: Being a finalist is the status; the order in which they are picked usually doesn't create a different outcome for "who are the finalists." This is a combination.
* Calculation: $C(15, 3) = \frac{15 \times 14 \times 13}{3 \times 2 \times 1}$.
* $15 / 3 = 5$.
* $14 / 2 = 7$.
* $5 \times 7 \times 13 = 35 \times 13 = 455$.
* Result: $455$

9. In how many ways can 6 students be seated at one side of a table with 4 chairs?
* Logic: There are more students than chairs. We need to pick 4 students out of 6 and seat them in specific chairs. Since seats are distinct positions, order matters. This is a permutation.
* Calculation: $P(6, 4) = 6 \times 5 \times 4 \times 3$.
* $30 \times 12 = 360$.
* Result: $360$

10. A traveler can choose from three airlines, five hotels, and four rental car companies. How many arrangements of these services are possible?
* Logic: The traveler needs one airline AND one hotel AND one car. We multiply the number of choices for each category.
* Calculation: $3 \times 5 \times 4$.
* $3 \times 5 = 15$.
* $15 \times 4 = 60$.
* Result: $60$

Final Answer:
1. 10,897,286,400
2. 12
3. 4,368
4. 21
5. 2,600
6. 126
7. 3,024
8. 455
9. 360
10. 60