Mathematics from zero
Combinatorics: multiple-choice review
Crux Multiple-choice synthesis across the combinatorics unit — the counting principle, permutations and factorials, combinations, and telling order-matters from order-does-not.
Your altitude — climbing toward senior
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You are at middle altitude — in the skySix questions that cut across the whole unit. Each one asks you to pick the right counting tool for a situation and run the number — the same decision you make whenever you count possibilities instead of listing them.
Goal
Confirm you can choose between the counting principle, permutations, and combinations, work the arithmetic, and explain why each rule multiplies or divides the way it does.
Quiz
Completed
A diner picks one of 5 mains, one of 4 sides, and one of 3 drinks. How many different meals are possible, and which rule applies?
Heads-up Adding answers 'how many menu items exist', not 'how many meals can I form'. Each main pairs with every side and every drink, so the counts multiply: 5 × 4 × 3 = 60.
Heads-up Every choice multiplies the possibilities; none is ignored. With 5 mains, 4 sides, and 3 drinks the count is 5 × 4 × 3 = 60.
Heads-up The number 60 is right but the reason is not. Nothing here is arranged in order from one set; these are three separate independent choices, which is the plain counting principle.
Quiz
Completed
In how many orders can 5 different books be arranged on a shelf?
Heads-up A placed book is used up and cannot reappear, so the choices shrink: 5 × 4 × 3 × 2 × 1, not 5 × 5. The answer is 5! = 120.
Heads-up Arranging on a shelf is exactly the case where order does matter — book A then B is a different shelf from B then A. This is a permutation, 5! = 120.
Heads-up Each position is an independent placement, so the counts multiply, not add: 5 × 4 × 3 × 2 × 1 = 120.
Quiz
Completed
A club awards a 1st-place and a 2nd-place prize among 8 members (no one wins twice). How many ways can the two prizes be handed out?
Heads-up Only 2 positions are filled, not all 8. Take just as many shrinking factors as positions: 8 × 7 = 56. The full 8! would order every member.
Heads-up 1st place and 2nd place are different roles, so order does matter — this is a permutation, not a combination. Do not divide by 2!; the answer is 8 × 7 = 56.
Heads-up The second prize cannot go to the first winner, so the second choice has 7 options, not 8 — and you multiply the positions' option counts: 8 × 7 = 56.
Quiz
Completed
A team of 3 is chosen from 6 candidates for one shared task (no roles). How many possible teams are there?
Heads-up Those 120 count each team once for every ordering of its 3 members. Since the team has no roles, order does not matter — divide by 3! = 6 to get 20.
Heads-up Choosing a team is not a single multiplication by the size. Count ordered picks 6 × 5 × 4 = 120, then divide by 3! = 6 because order does not matter, giving 20.
Heads-up A candidate cannot fill two slots, so picks shrink (6 × 5 × 4), and the slots are interchangeable, so divide by 3!. The answer is 120 ÷ 6 = 20.
Quiz
Completed
Two problems: (A) the number of ways to seat 4 people in a row, and (B) the number of ways to choose 4 of 4 people for an unordered group. How do the answers compare?
Heads-up Order distinguishes the two. Seating in a row cares about position (4! = 24); choosing an unordered group of all 4 gives just 1 — there is only one way to take everyone.
Heads-up A combination ignores order, so all 4! orderings of the same 4 people collapse into the single group: B = 24 ÷ 4! = 1.
Heads-up It is reversed. Arranging in a row (order matters) gives 4! = 24; choosing the unordered group of all 4 gives 1.
Quiz
Completed
You must pick 2 of 5 toppings for a pizza. Why is the answer 10 rather than the 20 you get from 5 × 4?
Heads-up You divide by the factorial of the group size you chose, which is 2! = 2 — not by the starting count. 20 ÷ 2 = 10.
Heads-up Repetition is not the issue; the two toppings are distinct. The 20 overcounts because order does not matter, and dividing by 2! corrects it to 10.
Heads-up Picking 2 toppings as an ordered list is 5 × 4 = 20 (the counts multiply). It overcounts only because order is irrelevant, so divide by 2! to get 10.
Recap
The whole unit reduces to one decision and one of two adjustments. First ask: are these separate choices, or am I drawing from one set? Separate independent choices — multiply their option counts (the counting principle). Drawing from one set — ask whether order matters. If it does, multiply the shrinking factors (a permutation; all of them is n!). If it does not, count the ordered picks and divide by r! to merge the duplicate orderings (a combination). Picking the wrong tool, or forgetting to divide out order, is what every distractor above was made of.