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: How many ways 5 runners can be arranged for 1st, 2nd, and 3rd place

This is about order mattering — who comes first, second, third matters. So this is a permutation.

We are choosing 3 people out of 5, and 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: Selecting 3 types of fruit from a basket of 10 different types of fruit

Here, we’re just picking fruits — no order matters. If you pick apple, banana, orange — it’s the same as orange, banana, apple. So this is a combination.

C(10, 3) = 10! / [3! × (10-3)!] = 10! / (3! × 7!)
= (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120

✔ Answer: Combination → 120 ways

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Problem 3: Choosing 4 books from a bin of 20 books

Again, unless specified that order matters (like arranging them on a shelf), “choosing” usually means combination.

So C(20, 4) = 20! / [4! × 16!]
= (20 × 19 × 18 × 17) / (4 × 3 × 2 × 1)
= (20×19×18×17) / 24
First: 20×19 = 380; 380×18 = 6840; 6840×17 = 116,280
Then: 116,280 ÷ 24 = 4,845

✔ Answer: Combination → 4,845 ways

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Problem 4: The batting order for a softball team of 20 players

Batting order = order matters → Permutation

But wait — how many batters? Usually 9 in a lineup. But the problem doesn’t say how many positions. Hmm.

Actually, re-reading: “the batting order for a softball team of 20 players” — this likely means we’re arranging ALL 20 players in order? That would be 20!.

But that seems too big. Maybe it’s implying we choose 9 batters from 20 and arrange them? But again, not specified.

Wait — let’s look at context. In most school problems, if they say “batting order for a team of X players”, they mean arranging all X players in order — so permutation of all.

But 20! is huge — maybe they meant choosing and arranging 9? Not clear.

Hold on — perhaps it’s a trick. Let me check standard interpretation.

Actually, in many textbooks, “batting order for a team of 20 players” implies selecting and ordering 9 players (since only 9 bat). But since it’s not specified, maybe we assume full arrangement?

Wait — looking back at Problem 1: “arranged for 1st, 2nd, 3rd” — specific positions.

Here, “batting order” typically means the sequence of batters — which is an ordered list. But how long? Since it says “for a softball team of 20 players”, I think it’s safe to assume we are arranging all 20 in order — but that’s unusual.

Alternatively, maybe it’s asking for number of possible batting orders using any subset? No.

I think there’s ambiguity, but in most similar problems, when they say “batting order for a team of N players”, they mean arranging all N players in order — so permutation of N items.

So P(20, 20) = 20! — but that’s astronomical.

Wait — perhaps it’s a typo or misphrasing. Maybe it’s “choosing 9 players from 20 and arranging them”? But not stated.

Looking at other problems — Problem 8 has electing officers from 300 students — that’s permutation because positions are distinct.

Similarly here — batting order has distinct positions (1st batter, 2nd, etc.). But how many positions? Softball has 9 batters.

I think the intended meaning is: how many ways to arrange 9 batters chosen from 20 players? That makes sense.

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 approximately 60.9 billion — but let's write it as 60,949,324,800

But maybe the problem expects us to recognize it’s a permutation without calculating? But others require calculation.

Wait — perhaps I overcomplicated. Let me see the original wording: “the batting order for a softball team of 20 players”

In some contexts, it might mean the number of ways to assign batting positions to all 20 players — but that doesn't make sense because only 9 bat.

I think the safest assumption is that we are to find the number of ways to choose and order 9 players from 20 for the batting lineup.

So yes, P(20,9) = 60,949,324,800

But let me confirm with calculator logic:

20P9 = 20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12

As above: 60,949,324,800

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

*(Note: If the problem meant something else, it should specify, but based on standard interpretation, this is correct.)*

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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.

So it’s 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

✔ Answer: 5040 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.)

Ah — “you can repeat toppings” — so this is with replacement, and also, does order matter? For pizza toppings, usually order doesn’t matter — pepperoni and mushroom is same as mushroom and pepperoni.

But since repeats are allowed, and order doesn’t matter, this is combinations with repetition.

Formula: 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) = (17160)/24 = 715

Wait: 13×12=156; 156×11=1716; 1716×10=17,160; divided by 24 = 715

✔ 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?

“Different types” and “combinations” — so order doesn’t matter, and no repeats (since different types).

So 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 — president ≠ VP ≠ secretary — so order matters. Also, probably no person can hold two offices, so no replacement.

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

Calculate:

300 × 299 = 89,700
89,700 × 298

Break it down: 89,700 × 300 = 26,910,000 minus 89,700 × 2 = 179,400
So 26,910,000 - 179,400 = 26,730,600

Wait — better: 89,700 × 298 = 89,700 × (300 - 2) = 89,700×300 - 89,700×2

89,700 × 300 = 26,910,000
89,700 × 2 = 179,400
Subtract: 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: A,B,C,D,E,F,G — that’s 7 distinct letters.

We want 4-letter arrangements, allow repeats, and order matters (since it says "arrangements").

So for each position, 7 choices.

Total = 7 × 7 × 7 × 7 = 7^4 = 2401

✔ Answer: 2401 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” — unless specified that positions matter, this is combination.

So 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

So 17,100,720 ÷ 120

Divide: 17,100,720 ÷ 120 = 17,100,720 ÷ 12 ÷ 10 = first ÷10 = 1,710,072; then ÷12

1,710,072 ÷ 12:

12 × 142,506 = ? Let me compute:

12 × 140,000 = 1,680,000
Subtract: 1,710,072 - 1,680,000 = 30,072
12 × 2,506 = 12×2500=30,000; 12×6=72 → total 30,072
So 140,000 + 2,506 = 142,506

Thus, 142,506

✔ Answer: 142,506 ways

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Now, compiling all answers:

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

Final Answer:
1. Permutation, 60
2. Combination, 120
3. Combination, 4845
4. Permutation, 60949324800
5. 5040
6. 715
7. 45
8. 26730600
9. 2401
10. 142506