Core Ideas in Plain Terms

Study Sheet

Core Ideas in Plain Terms

Counting and chance, in everyday language

The Counting Principle

Concept
Multiply independent choices
3 rows of 4

When you make several choices that don't affect each other, multiply the number of options. When choices are alternatives (this or that), add them. This is the Fundamental Counting Principle, and it underlies almost all counting.

Example
Outfits
3 rows of 4

With 33 shirts and 44 pairs of pants, the Fundamental Counting Principle gives 3×4=123 \times 4 = 12 outfits.

Arrangements vs. Selections

Concept
Does order matter?

If rearranging counts as different (a race finish, a seating), use permutations: nPk=n!(n−k)!{}_nP_k = \tfrac{n!}{(n-k)!}. If order does not matter (a committee, a handful of cards), use combinations: (nk)=n!k!(n−k)!\binom nk = \tfrac{n!}{k!(n-k)!}. The only difference is dividing out the rearrangements.

Example
Choose vs. arrange

Arranging 33 of 88 people in a row: 8P3=336{}_8P_3 = 336. Choosing a committee of 33: (83)=56\binom 83 = 56.

Probability

Concept
Favorable over total

For equally likely outcomes, the probability of an event is given by the classical probability formula P(A)=favorable outcomestotal outcomesP(A) = \dfrac{\text{favorable outcomes}}{\text{total outcomes}}. For an "at least one" question, it is usually easier to find the probability it doesn't happen and subtract from 11.

Example
At least one head

In 33 coin flips, the chance of no heads is (12)3=18\left(\tfrac12\right)^3 = \tfrac18, so at least one head has probability 1−18=781 - \tfrac18 = \tfrac78.

Going Deeper: Complements and Expected Value

Concept
The complement shortcut

The complement rule says P(A)=1−P(not A)P(A) = 1 - P(\text{not } A). That makes "at least one" problems almost always easier backwards: P(at least one)=1−P(none)P(\text{at least one}) = 1 - P(\text{none}). Roll a die 44 times — P(at least one six)=1−(56)4=1−6251296=6711296P(\text{at least one six}) = 1 - \left(\tfrac56\right)^4 = 1 - \tfrac{625}{1296} = \tfrac{671}{1296}, one subtraction instead of four cases.

Concept
Expected value is the long-run average

The expected value formula is E=∑(outcome×probability)E = \sum \big(\text{outcome} \times \text{probability}\big): multiply each outcome by its probability and add. A fair die averages 1+2+3+4+5+66=3.5\dfrac{1+2+3+4+5+6}{6} = 3.5 per roll — never an actual roll, but exactly what the average approaches over many rolls.

Example
Pigeonhole in one line

Any 1313 people include two born in the same month: 1313 pigeons, 1212 holes — some hole gets two. No probability needed, just counting.

Example
A committee count (multi-step)

From 55 teachers and 88 students, choose a committee of 22 teachers and 33 students.

Step 1 — teachers: (52)=10\binom{5}{2} = 10.

Step 2 — students: (83)=56\binom{8}{3} = 56.

Step 3 — the choices are independent, so by the Fundamental Counting Principle: 10×56=56010 \times 56 = 560 committees.

Problem-Solving Playbook

Concept
Direct, cases, or complement?

Three ways to count, in order of preference: directly (multiply choices), by cases (split cleanly, add), or by complement (count what you don't want, subtract). If identical arrangements got counted more than once, divide out the repeats.

Example
Worked: complement counting

Choose 44 of 66 people for a committee, but two of them refuse to serve together. Total: (64)=15\binom{6}{4} = 15. Committees containing BOTH: the other 22 seats from the remaining 44 people, (42)=6\binom{4}{2} = 6. By the complement rule: 15−6=915 - 6 = 9 committees.