Unit 1 Practice Problems

Challenge yourself with these counting problems. Try to solve them before looking at the hints!

Problem 1: Distributing Assignments

Permutations

Suppose there are 4 people and 9 different assignments. Each person should receive exactly one assignment. Assignments for different people must be different. How many ways are there to do it?

Ask yourself:

  • Are the items (assignments) distinct? Yes.
  • Does order matter (who gets what)? Yes.
  • Is repetition allowed? No (assignments must be different).

Conclusion: This is a Permutation problem (Ordered, No Repetition).

We are choosing 4 assignments out of 9, and order matters.

Formula: \(P(n, k) = \frac{n!}{(n-k)!}\)

Calculation: \(9 \times 8 \times 7 \times 6\)

3,024 ways

Problem 2: Distributing Candies

Stars & Bars

There are 15 identical candies. How many ways are there to distribute them among 7 kids?

  • Candies are identical → Order doesn't matter.
  • Kids are distinct → We are assigning counts to kids.
  • Repetition allowed? Yes, a kid can get multiple candies.

Conclusion: Combinations with Repetition (Stars and Bars).

\(n = 7\) (kids/bins), \(k = 15\) (candies/stars)

Formula: \(\binom{n+k-1}{k} = \binom{n+k-1}{n-1}\)

Calculation: \(\binom{15+7-1}{7-1} = \binom{21}{6}\)

54,264 ways

Problem 3: Fair Candy Distribution

Advanced Stars & Bars

Same as Problem 2 (15 candies, 7 kids), but now each kid must receive at least 1 candy.

Give 1 candy to each kid first to satisfy the condition.

Candies used: 7.

Candies remaining: \(15 - 7 = 8\).

Now distribute the remaining 8 candies among 7 kids with no restrictions.

\(n = 7, k = 8\)

Formula: \(\binom{8+7-1}{7-1} = \binom{14}{6}\)

3,003 ways

Problem 4: Working Groups

Complex Counting

There are 12 students. How many ways can you split them into 6 working groups of size 2?

Pick 2 for group 1: \(\binom{12}{2}\)

Pick 2 for group 2: \(\binom{10}{2}\)

...and so on until \(\binom{2}{2}\).

Product: \(\binom{12}{2} \times \binom{10}{2} \times \dots \times \binom{2}{2}\)

Wait! The order of groups doesn't matter.

We counted the sequence of groups (Group A, then Group B...) but {A, B} is the same set of groups as {B, A}.

Since there are 6 groups, we must divide by \(6!\).

\(\frac{12!}{2^6 \times 6!} =\)

10,395 ways

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