Permutations and Combinations

The shortest route to P&C competence is one diagnostic question asked relentlessly: does order matter? Everything in this topic, typically 4–8 marks per session, often on Paper 2, hangs off that test and the multiplication principle.

The counting principle

If task one has $m$ outcomes and task two has $n$, the pair has $m \times n$. Everything else in the topic is this principle dressed up. A four-digit code from digits 1–9 without repeats: $9 \times 8 \times 7 \times 6 = 3024$. Writing the slot-by-slot product is both the clearest thinking tool and the visible method the mark scheme rewards.

Factorials, nPr, nCr

• $n!$, arrangements of $n$ distinct items in a row
• ${}^nP_r = n!/(n - r)!$, ordered selections: $r$ items from $n$ where order matters
• ${}^nC_r = n!/(r!(n - r)!)$, unordered selections: $r$ items from $n$ where order doesn’t

The relationship ${}^nP_r = {}^nC_r \times r!$ says it cleanly: permutations are combinations then arranged. Both formulas are on the memorise list; calculator buttons exist on Paper 2, but the slot method keeps you honest about what you’re computing.

The order test in practice: arranging books on a shelf, forming numbers or “words”, seating rows, distinct prizes $\to$ order matters (${}^nP_r$ / factorials / slots). Committees, teams, selections of selections $\to$ order doesn’t (${}^nC_r$).

Restrictions: handle them first

The exam’s favourite escalation. The rule: place the constrained element first, then count the remaining freedoms.

How many 4-digit even numbers can be formed from 1, 2, 3, 4, 5, 6, 7 without repetition? Last digit even: 3 choices (2, 4, 6). Then first three digits from the remaining 6: $6 \times 5 \times 4 = 120$. Total $= 120 \times 3 =$ 360

“Together” constraints: glue the items into one block (count it as a unit, times the arrangements within the block). “Separated” constraints: total minus together, the complement.

5 students in a row, two particular ones together: treat the pair as a block $\to 4! \times 2! = 48$ The same two NOT together: $5! - 48 = 120 - 48 =$ 72

”At least” questions: count the complement

“At least one woman on the 4-person committee from 6 men, 5 women”: all-men committees are the only failures $\to {}^{11}C_4 - {}^6C_4 = 330 - 15 =$ 315. One subtraction versus four added cases, and the four-case route invites both slips and double counting. State the logic in a few words (“total $-$ no women”); the stated structure is method credit.

Worked exam-style question

A team of 5 is chosen from 6 boys and 4 girls. Find the number of teams with more boys than girls.

More boys than girls in 5 picks: 3B+2G, 4B+1G, 5B+0G (B, cases identified) 3B2G: ${}^6C_3 \times {}^4C_2 = 20 \times 6 = 120$ (M) 4B1G: ${}^6C_4 \times {}^4C_1 = 15 \times 4 = 60$ 5B0G: ${}^6C_5 = 6$ Total $= 120 + 60 + 6 =$ 186 (A)

The structure, cases listed, each a product of combinations, then summed, is the entire method. Cases that overlap or omit are where this topic’s marks die; writing the case list first prevents both.

Common mistakes in this topic

• ${}^nP_r$/${}^nC_r$ chosen by vibe instead of the order test
• Restrictions applied after free counting (double counts)
• “Together” blocks counted without the internal $\times 2!$ (or $\times k!$)
• “At least” computed case-by-case with a case missing
• Answers left as formulas: ${}^6C_3$ written but never evaluated

This topic shares its ${}^nC_r$ machinery with binomial expansion in series, so the drilling transfers. It’s also a classically avoided topic, which makes it a cheap differentiator at the A/A* boundary.

P&C confusion is nearly always one missing question. “does order matter?”, asked at the right moment. One session fixes the habit: free 1-hour trial with Teacher Rig via WhatsApp.