Permutations and Combinations Notes and Worksheets - Lindsay Bowden - Free Printable

Let’s go through each problem one by one. We’ll figure out whether it’s a permutation or combination, and then calculate the answer if needed.

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**Problem 1: Permutation or Combination?
How many ways 5 runners can be arranged for 1st, 2nd, and 3rd place**

→ Order matters here (1st, 2nd, 3rd are different positions). So this is a permutation.

We’re choosing 3 people out of 5, where order matters.

Formula: P(n, r) = n! / (n - r)!

P(5, 3) = 5! / (5-3)! = 5! / 2! = (5 × 4 × 3 × 2 × 1) / (2 × 1) = 120 / 2 = 60

✔ Answer: Permutation, 60 ways

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**Problem 2: Permutation or Combination?
Selecting 3 types of fruit from a basket of 10 different types of fruit**

→ You’re just picking 3 fruits — no order matters. So this is a combination.

C(10, 3) = 10! / [3! × (10-3)!] = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120

✔ Answer: Combination, 120 ways

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**Problem 3: Permutation or Combination?
Choosing 4 books from a bin of 20 books**

→ Just selecting books — no order mentioned. So combination.

C(20, 4) = 20! / [4! × 16!] = (20 × 19 × 18 × 17) / (4 × 3 × 2 × 1)

Calculate numerator: 20×19=380, 380×18=6840, 6840×17=116,280

Denominator: 24

116,280 ÷ 24 = 4,845

✔ Answer: Combination, 4,845 ways

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**Problem 4: Permutation or Combination?
The batting order for a softball team of 20 players**

→ Batting order means sequence matters → permutation

But wait — how many batters? The question doesn’t say how many are in the lineup. In softball, usually 9 batters. But since it says “batting order for a softball team of 20 players”, it might mean arranging all 20? That seems odd.

Actually, re-reading: “the batting order” implies we’re ordering some number of players. Since it doesn’t specify, but typically you bat 9 players, but the problem says “team of 20 players” — maybe it’s asking for how many ways to arrange the entire team’s batting order? That would be 20!.

But that’s huge. Let me check context — other problems use small numbers. Maybe it’s implied we’re choosing and ordering 9? But the problem doesn’t say.

Wait — looking again: “the batting order for a softball team of 20 players” — I think it means we are creating an ordered list of all 20 players’ batting positions. So yes, permutation of all 20.

So P(20, 20) = 20! — which is enormous. But perhaps they mean choosing 9 out of 20 and ordering them? The problem is ambiguous.

But in standard math problems, when they say “batting order for a team of X players”, they often mean arranging all X in order. However, in reality, only 9 bat. But since the problem doesn’t specify, and given the context of other problems, let’s assume it’s arranging all 20 in order.

But 20! is too big — maybe it’s a trick? Or perhaps it’s choosing 9?

Looking at Problem 8: electing 3 officers from 300 — so they do specify numbers.

Here, it just says “batting order for a softball team of 20 players”. I think safest assumption is that we are ordering all 20 players in a line — so permutation of 20 items.

But that’s 20! = 2,432,902,008,176,640,000 — too big for a worksheet.

Alternative interpretation: perhaps it’s how many ways to choose and order 9 batters from 20? That makes more sense.

In many textbooks, “batting order” implies selecting and ordering 9 players from the team.

I think that’s intended. Let’s go with that.

So P(20, 9) = 20! / (20-9)! = 20! / 11!

That’s still big, but calculable:

20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12

Let me compute step by step:

20×19 = 380

380×18 = 6,840

6,840×17 = 116,280

116,280×16 = 1,860,480

1,860,480×15 = 27,907,200

27,907,200×14 = 390,700,800

390,700,800×13 = 5,079,110,400

5,079,110,400×12 = 60,949,324,800

So 60,949,324,800

But that’s huge. Maybe the problem means something else?

Another thought: perhaps “batting order” means just the sequence, and we’re to find how many possible orders for the whole team — but again, 20! is astronomical.

Perhaps it’s a typo, and it’s supposed to be 9 players? But it says 20.

Looking back at the image — it says “softball team of 20 players” — and in context, other problems have reasonable answers. Problem 8 has 300 students choosing 3 officers — which is P(300,3) = 300×299×298 = about 26 million — also big but manageable.

P(20,9) is 60 billion — very big.

Perhaps it’s combination? No, order matters in batting order.

I think we have to go with P(20,9) as the intended meaning.

But let’s see Problem 5: arranging 7 numbers — that’s 7! = 5040, reasonable.

Problem 6: pizza toppings with repetition — 10^4 = 10,000.

Problem 7: C(10,2) = 45.

Problem 8: P(300,3) = 300*299*298.

Let me calculate that: 300*299 = 89,700; 89,700*298.

89,700*300 = 26,910,000 minus 89,700*2 = 179,400 → 26,910,000 - 179,400 = 26,730,600.

So around 26 million.

P(20,9) is 60 billion — much bigger. Perhaps it's not that.

Another idea: maybe "batting order" means we are to find how many ways to arrange the 20 players in a line for batting — i.e., 20!.

But 20! is 2.4e18 — way too big.

Perhaps the problem is misstated, and it's supposed to be "choosing 9 players for the batting order" or something.

Given the context, and to match the scale of other problems, I think it's likely that they mean selecting and ordering 9 players from 20.

I'll proceed with P(20,9) = 60,949,324,800.

But let's double-check online or standard problems — upon second thought, in many curricula, when they say "batting order for a team of n players", they mean the number of ways to order all n players, but that's only if n is small. Here n=20 is large.

Perhaps it's a combination? No.

Another possibility: maybe "batting order" refers to the sequence, but they want the number of possible lineups, which is P(20,9).

I think we have to go with that.

So ✔ Answer: Permutation, 60,949,324,800 ways

But let's write it as 6.09493248 × 10^10, but probably leave as is.

For now, I'll keep it.

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Problem 5: How many ways can you arrange these 7 numbers: 1,2,3,5,7,8,9? (Don't repeat.)

→ Arranging all 7 distinct numbers — order matters, no repetition.

This is 7! = 7×6×5×4×3×2×1 = 5040

✔ Answer: 5,040 ways

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Problem 6: You are choosing 4 pizza toppings from a menu of 10 toppings. How many different pizzas can you make? (You can repeat toppings.)

→ Repetition allowed, and order doesn't matter for pizza toppings (usually, unless specified).

The problem says "you can repeat toppings", and it's about making pizzas, so likely order doesn't matter — e.g., pepperoni and mushroom is same as mushroom and pepperoni.

But when repetition is allowed and order doesn't matter, it's combinations with repetition.

Formula for combinations with repetition: C(n + r - 1, r) = C(10 + 4 - 1, 4) = C(13,4)

C(13,4) = 13! / (4! * 9!) = (13×12×11×10)/(4×3×2×1) = 17,160 / 24 = 715

Is that correct? Let me confirm.

Yes, for choosing r items from n types with repetition allowed and order not important, it's C(n+r-1, r).

So C(13,4) = 715.

If order mattered, it would be 10^4 = 10,000, but for pizza, order usually doesn't matter.

The problem says "different pizzas", and typically toppings are unordered.

So I think 715 is correct.

✔ Answer: 715 different pizzas

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Problem 7: Joseph wants 2 different types of soda for a party. There are 10 different types of soda at the store. How many different combinations of soda could he choose?

→ Choosing 2 different sodas, order doesn't matter (combinations).

C(10,2) = 10! / (2! * 8!) = (10×9)/2 = 45

✔ Answer: 45 combinations

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Problem 8: The student body is electing class officers for president, vice president, and secretary. If there are 300 students, how many different possible selections are there?

→ Positions are distinct, so order matters. Also, presumably one person per position, no repeats.

So permutation: P(300,3) = 300 × 299 × 298

Calculate:

300 × 299 = 89,700

89,700 × 298

First, 89,700 × 300 = 26,910,000

Minus 89,700 × 2 = 179,400

So 26,910,000 - 179,400 = 26,730,600

✔ Answer: 26,730,600 possible selections

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Problem 9: How many different 4 letter arrangements can you make with the letters ABCDEFG? (Allow repeats.)

→ Letters are A,B,C,D,E,F,G — 7 distinct letters.

Arrangements of 4 letters, with repetition allowed, and order matters (since it says "arrangements").

So for each position, 7 choices.

Total = 7^4 = 7×7×7×7 = 49×49 = 2,401

✔ Answer: 2,401 arrangements

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Problem 10: Find the number of possible outcomes from choosing 5 basketball players from a group of 30 that tried out.

→ Choosing 5 players, no mention of order, so combination.

C(30,5) = 30! / (5! * 25!) = (30×29×28×27×26) / (5×4×3×2×1)

Numerator: 30×29=870, 870×28=24,360, 24,360×27=657,720, 657,720×26=17,100,720

Denominator: 120

17,100,720 ÷ 120 = ?

First, 17,100,720 ÷ 10 = 1,710,072

Then ÷ 12 = 1,710,072 ÷ 12

12 × 142,506 = 12×140,000=1,680,000; 12×2,506=30,072; total 1,680,000+30,072=1,710,072

So 142,506

Wait, let me compute directly:

30×29×28×27×26 / 120

Note that 30/5 = 6, so better to simplify:

(30/5) × 29 × (28/4) × (27/3) × (26/2) / (1) wait no.

Standard way:

= [30×29×28×27×26] / [5×4×3×2×1]

Break down:

30/5 = 6

28/4 = 7

27/3 = 9

26/2 = 13

And 29 remains.

Denominator left: 1 (since 5,4,3,2 used)

After dividing:

We have 6 × 29 × 7 × 9 × 13

Now compute step by step:

6×29 = 174

174×7 = 1,218

1,218×9 = 10,962

10,962×13 = 10,962×10 = 109,620; 10,962×3=32,886; total 109,620+32,886=142,506

Yes.

✔ Answer: 142,506 ways

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Now, summarizing all answers:

1. Permutation, 60
2. Combination, 120
3. Combination, 4,845
4. Permutation, 60,949,324,800 (assuming P(20,9))
5. 5,040
6. 715
7. 45
8. 26,730,600
9. 2,401
10. 142,506

For problem 4, I'm still unsure. Let me think again.

In some contexts, "batting order" might mean the number of ways to arrange the entire team, but that's 20! which is too big.

Perhaps it's a combination? No.

Another idea: maybe "for a softball team of 20 players" means we have 20 players, and we need to choose who bats in what order, but typically only 9 bat, so P(20,9).

I think that's standard.

Perhaps the problem is from a source where they expect P(20,9).

I'll stick with that.

Final Answers:

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Final Answer:
1. Permutation, 60
2. Combination, 120
3. Combination, 4,845
4. Permutation, 60,949,324,800
5. 5,040
6. 715
7. 45
8. 26,730,600
9. 2,401
10. 142,506