Permutations And Combinations Worksheet Answers Lovely ... - Free Printable

Let’s solve each problem step by step and match it to the correct answer letter.

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Problem 1: How many different ways can you have 1st, 2nd, 3rd, and 4th in a race with 10 runners?

→ This is about order matters — who comes first, second, etc. So we use permutations.

We choose 4 runners out of 10 and arrange them in order:
P(10,4) = 10 × 9 × 8 × 7 = 5040

✔ Matches → d. 5040

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Problem 2: How many different combinations of management can there be to fill the positions of manager, secretary, and advisor from 12 eligible candidates?

→ Again, positions are distinct (manager ≠ secretary ≠ advisor), so order matters → permutations.

Choose 3 people from 12 and assign to specific roles:
P(12,3) = 12 × 11 × 10 = 1320

✔ Matches → h. 1320

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Problem 3: How many different words can we make using the letters S, H, A, and L? (Assume every combination is a word.)

→ We’re arranging all 4 distinct letters → 4! = 4 × 3 × 2 × 1 = 24

✔ Matches → g. 24

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Problem 4: Robin has 5 different pairs of shoes and 6 different pairs of socks. How many shoe-and-sock combinations if he picks one pair of each?

→ For each shoe choice, he can pick any sock → multiplication principle.

5 × 6 = 30

✔ Matches → b. 30

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Problem 5: Ginny has 6 water colors. In how many different orders can they appear?

→ Arranging all 6 distinct items → 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

✔ Matches → f. 720

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Problem 6: How many different sandwiches can be made with 6 breads, 4 cheeses, and 3 potatoes? One from each category.

→ Multiply choices: 6 × 4 × 3 = 72

✔ Matches → c. 72

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Problem 7: How many 2-digit numbers can you make using digits 1, 2, 3, 4 without repeating digits?

→ First digit: 4 choices
Second digit: 3 remaining choices (no repeat)
Total: 4 × 3 = 12

✔ Matches → a. 12

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Problem 8: Eva has 8 glasses of water. In how many different orders can they be arranged?

→ Assuming all glasses are identical? Wait — but the question says “different orders”, which implies we’re treating them as distinguishable (maybe labeled or just counting arrangements).

If all 8 are distinct → 8! = 40320

But let’s check: 8! = 40320 → matches option e

Wait — is that right? Let me double-check:

8! = 8×7×6×5×4×3×2×1 = 40320 ✔

So yes → e. 40320

BUT — hold on! The worksheet lists only up to h=1320, and e=40320 is listed. So yes, it fits.

Wait — actually, looking back at the options:

a. 12
b. 30
c. 72
d. 5040
e. 40320 ← this is 8!
f. 720 ← 6!
g. 24 ← 4!
h. 1320 ← P(12,3)

Yes, so Problem 8 → 8! = 40320 → e

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Now let’s list all matches:

1 → d
2 → h
3 → g
4 → b
5 → f
6 → c
7 → a
8 → e

Final Answer:

1-d, 2-h, 3-g, 4-b, 5-f, 6-c, 7-a, 8-e