Counting, combinations and permutations

1. There are 3 stages: rice, protein, salsa.
2. By the fundamental counting principle, multiply the number of choices at each stage: 2 \times 5 \times 4 = 40
3. So 40 different burrito bowls can be made.

Check: For each of the 2 kinds of rice there are 5 \times 4 = 20 protein-and-salsa pairs. 2 \times 20 = 40 ✓

Tip: "One of each" from separate lists means multiply, never add. Adding (2 + 5 + 4 = 11) only counts the items, not the combinations of them.

1. Putting all 6 presentations in order: order matters and no student presents twice.
2. 1st spot: 6 choices. 2nd: 5. 3rd: 4. 4th: 3. 5th: 2. 6th: 1.
3. Multiply: 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
4. There are 720 different orders.

Check: 6 \times 5 = 30, 30 \times 4 = 120, 120 \times 3 = 360, 360 \times 2 = 720, 720 \times 1 = 720 ✓

Tip: Arranging all n different items in a row is always n!. On the TI-30XS: 6, prb, choose !, enter.

1. Building an outfit has 3 stages: shirt, pants, shoes.
2. The fundamental counting principle says to multiply the number of choices at each stage.
3. 5 \times 3 \times 4 = 60
4. Dana can make 60 different outfits.

Check: Each shirt goes with 3 \times 4 = 12 pants-and-shoes pairs. Five shirts give 5 \times 12 = 60 ✓

Tip: When you pick one item from each of several separate lists, multiply the list sizes. Adding (5 + 3 + 4 = 12) counts the pieces of clothing in the suitcase, not the outfits.

1. The offices have different titles, so order matters: Ana as president and Ben as treasurer is different from the reverse. This is a permutation.
2. President: 11 choices. Secretary: 10 left. Treasurer: 9 left.
3. Multiply: {}_{11}P_3 = 11 \times 10 \times 9 = 990
4. The offices can be filled in 990 ways.

Check: Using the formula, \dfrac{11!}{(11-3)!} = \dfrac{11!}{8!} = 11 \times 10 \times 9 = 990 ✓

Tip: Different titles or positions (1st/2nd/3rd, president/secretary) mean order matters — use a permutation. Only use a combination when the chosen people are all equal.

1. The order of the toppings doesn't change the cup, so order does not matter → combination.
2. Use {}_nC_r = \dfrac{{}_nP_r}{r!} with n = 9, r = 3.
3. Permutations: 9 \times 8 \times 7 = 504.
4. Each group of 3 toppings can be ordered in 3! = 6 ways, so divide: {}_9C_3 = \frac{504}{6} = 84
5. There are 84 different cups.

Check: On the TI-30XS, 9 nCr 3 gives 84 ✓

Tip: When order doesn't matter, first count as if it did, then divide by r! to remove the duplicate orderings.

1. The drink is one item: a hot drink or a juice. "Or" inside one stage means add: 5 + 2 = 7 \text{ drinks}
2. Then choose a drink and a main dish. "And" between stages means multiply: 7 \times 6 = 42
3. There are 42 breakfast combos.

Check: Hot-drink combos: 5 \times 6 = 30. Juice combos: 2 \times 6 = 12. Total 30 + 12 = 42 ✓

Tip: Multiply between stages ("and"); add the options inside one stage ("or").

1. There are 5 questions, and each one is a stage.
2. Each question has 4 possible answers, and the same letter can be used again on another question (repeats are allowed).
3. Multiply 4 by itself 5 times: 4 \times 4 \times 4 \times 4 \times 4 = 4^5 = 1{,}024
4. There are 1,024 different answer sheets.

Check: 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1{,}024 ✓

Tip: With repeats allowed, use (\text{choices per position})^{(\text{number of positions})}. The base is 4 (choices) and the exponent is 5 (questions) — swapping them gives 5^4 = 625, a common mistake.

1. The awards are different, so order matters: winning Best Film is not the same as winning Third Place. This is a permutation.
2. Best Film: 9 choices. Second Place: 8 films left. Third Place: 7 films left.
3. Multiply: {}_9P_3 = 9 \times 8 \times 7 = 504
4. The awards can be given in 504 ways.

Check: {}_9C_3 = 84 groups of three films, and each group can be ranked in 3! = 6 ways: 84 \times 6 = 504 ✓

Tip: Ranked places (1st, 2nd, 3rd) always mean order matters. Count down from n, one factor per place.

1. Only the group of books matters, not the order, so this is a combination.
2. First count ordered choices: 9 \times 8 \times 7 \times 6 = 3{,}024.
3. Each group of 4 books can be put in order in 4! = 4 \times 3 \times 2 \times 1 = 24 ways, so divide: {}_9C_4 = \frac{3{,}024}{24} = 126
4. Jordan can choose 126 different groups.

Check: On the TI-30XS, 9, prb, choose nCr, 4, enter gives 126 ✓. Also {}_9C_4 = {}_9C_5 (choosing 4 to read is the same as choosing 5 to skip) ✓

Tip: For a combination, multiply r numbers counting down from n, then divide by r!.

1. The phone is one choice: an Android or an iPhone. Inside a single stage, "or" means add: 3 + 2 = 5 \text{ phones}
2. Then pick a phone and a data plan. Between stages, "and" means multiply: 5 \times 4 = 20
3. There are 20 different deals.

Check: Android deals: 3 \times 4 = 12. iPhone deals: 2 \times 4 = 8. Total 12 + 8 = 20 ✓

Tip: You get only one phone, so the phone groups are added together before you multiply by the next stage.

1. Members have equal roles, so use combinations for each group.
2. Teachers: {}_6C_2 = \frac{6 \times 5}{2 \times 1} = 15
3. Parents: {}_8C_2 = \frac{8 \times 7}{2 \times 1} = 28
4. Any teacher pair can serve with any parent pair, so multiply: 15 \times 28 = 420
5. There are 420 possible committees.

Check: 15 \times 28 = 15 \times 30 - 15 \times 2 = 450 - 30 = 420 ✓

Tip: "2 from this group and 2 from that group" → find each combination separately, then multiply. Using {}_{14}C_4 = 1{,}001 would allow committees with 4 teachers and no parents.

1. A route is an order of all 6 houses, and each house is visited once. That is an arrangement of all 6 items: 6!.
2. First stop: 6 choices; second: 5; third: 4; fourth: 3; fifth: 2; last: 1.
3. 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
4. There are 720 possible routes.

Check: 5! = 120 and 6! = 6 \times 5! = 6 \times 120 = 720 ✓

Tip: Each new factorial is the previous one times the next number: 5! = 120, so 6! = 720 and 7! = 5{,}040. Knowing these saves time.

1. Fill the restricted seat first. Rosa must be in the middle chair: 1 choice.
2. The other 4 coworkers fill the remaining 4 chairs in any order: 4! = 4 \times 3 \times 2 \times 1 = 24
3. Multiply: 1 \times 24 = 24.
4. There are 24 arrangements.

Check: Without the rule there are 5! = 120 arrangements. Rosa is in each of the 5 chairs equally often, so \tfrac{1}{5} of them have her in the middle: 120 \div 5 = 24 ✓

Tip: A person or item locked into one spot has just 1 choice. Place it first, then arrange everything else.

1. The tree has 2 \times 3 \times 2 = 12 end branches (tickets) in all.
2. Follow only the Train branch. It splits into Morning, Afternoon and Evening.
3. Keep only Afternoon and Evening: 2 departure times.
4. Each of those splits into One-way and Round-trip: 2 trip types.
5. Count the end branches: 1 \times 2 \times 2 = 4 (Train–Afternoon–One-way, Train–Afternoon–Round-trip, Train–Evening–One-way, Train–Evening–Round-trip).
6. 4 tickets fit.

Check: Train has 3 \times 2 = 6 tickets; removing the 2 morning tickets leaves 6 - 2 = 4 ✓

Tip: A tree diagram is the counting principle drawn out. To count outcomes with a condition, follow only the branches that meet it and multiply what's left.

1. A number is even when its last digit is even. Fill this restricted position first.
2. Last digit: must be 2, 4, 6 or 8 → 4 choices.
3. First digit: any of 1–9 except the one already used → 8 choices.
4. Second digit: any digit not yet used → 7 choices.
5. Third digit: any digit not yet used → 6 choices.
6. Multiply: 4 \times 8 \times 7 \times 6 = 1{,}344
7. There are 1,344 codes.

Check: Without the "even" rule there would be 9 \times 8 \times 7 \times 6 = 3{,}024 codes. Four of the nine digits are even, so \tfrac{4}{9} of the codes should be even: 3{,}024 \times \tfrac{4}{9} = 1{,}344 ✓

Tip: Always fill the most restricted position first, then count the remaining positions.

1. The order doesn't matter and numbers can't repeat, so this is a combination: {}_{20}C_5.
2. Ordered count: 20 \times 19 \times 18 \times 17 \times 16 = 1{,}860{,}480.
3. Each set of 5 numbers can be ordered in 5! = 120 ways, so divide: \frac{1{,}860{,}480}{120} = 15{,}504
4. There are 15,504 possible sets.

Check: On the TI-30XS, 20 nCr 5 = 15,504 ✓. Also 15{,}504 \times 120 = 1{,}860{,}480 ✓

Tip: Lottery picks, committees, and hands of cards are combinations. Only if the numbers had to match in the exact order would it be a permutation.

1. Fill the restricted position first. First digit: any of 1–9 → 9 choices.
2. Second digit: any of the 10 digits except the one already used. 0 is now allowed, so 10 - 1 = 9 choices.
3. Third digit: 10 - 2 = 8 choices.
4. Fourth digit: 10 - 3 = 7 choices.
5. Multiply: 9 \times 9 \times 8 \times 7 = 4{,}536
6. There are 4,536 codes.

Check: All no-repeat codes: 10 \times 9 \times 8 \times 7 = 5{,}040. Codes starting with 0: 1 \times 9 \times 8 \times 7 = 504. 5{,}040 - 504 = 4{,}536 ✓

Tip: The second position gets 0 back as an option. That is why it has 9 choices, not 8.

1. Choosing 2 people where order doesn't matter: {}_nC_2 = \frac{n(n-1)}{2}
2. Set it equal to 45: \frac{n(n-1)}{2} = 45 \quad\Rightarrow\quad n(n-1) = 90
3. Look for two consecutive whole numbers whose product is 90: 10 \times 9 = 90.
4. So n = 10. The team has 10 employees.

Check: {}_{10}C_2 = \dfrac{10 \times 9}{2} = \dfrac{90}{2} = 45 ✓

Tip: In a "work backwards" counting problem, plugging each answer choice into the formula is often the fastest method. Test 9: \frac{9 \times 8}{2} = 36 ✗. Test 10: 45 ✓.

1. The offices are different, so order matters: {}_nP_2 = n(n-1).
2. Set it equal to the number of ways: n(n-1) = 132
3. Find two consecutive whole numbers whose product is 132. Since \sqrt{132} \approx 11.5, try 12 and 11: 12 \times 11 = 132 ✓
4. So n = 12. The club has 12 members.

Check: 12 choices for president, then 11 for vice-president: 12 \times 11 = 132 ✓

Tip: For n(n-1) = N, the answer is just above \sqrt{N}. Testing the choices is also quick: 11 \times 10 = 110 ✗, 12 \times 11 = 132 ✓.

1. Rule A: 4 positions, 26 choices each: 26^4 = 26 \times 26 \times 26 \times 26 = 456{,}976
2. Rule B: 6 positions, 10 choices each: 10^6 = 1{,}000{,}000
3. Compare: 1{,}000{,}000 > 456{,}976, so Rule B allows more.
4. Difference: 1{,}000{,}000 - 456{,}976 = 543{,}024.
5. Rule B, by 543,024 passwords.

Check: 26^2 = 676 and 676^2 = 456{,}976 ✓; 456{,}976 + 543{,}024 = 1{,}000{,}000 ✓

Tip: More positions can beat more choices per position. Always compute both counts — don't judge by the bigger alphabet alone.

1. Now: 4 \times 6 \times 3 = 72 burritos.
2. New tortilla: 5 \times 6 \times 3 = 90 → an increase of 90 - 72 = 18.
3. New filling: 4 \times 7 \times 3 = 84 → an increase of 84 - 72 = 12.
4. New salsa: 4 \times 6 \times 4 = 96 → an increase of 96 - 72 = 24.
5. The largest increase comes from a new salsa (+24).

Check: A new item in one list pairs with every combination from the other two lists. New salsa: 4 \times 6 = 24 new burritos ✓; new tortilla: 6 \times 3 = 18 ✓; new filling: 4 \times 3 = 12 ✓

Tip: Adding one item to the smallest list helps most, because that item pairs with the product of the larger lists.