PROBABILITY · HL
Combinations
Order doesn't matter.
Section 1 of 11
Order is unimportant
A combination is a selection where the order doesn't matter.
If you pick A then B, or B then A, it's the same pair. Counted once.
Compare with permutations — ${}^{n}P_{r}$ — where order does matter.
(i) Worked example — 2 from 4
How many ways can 2 boys be chosen from a group of 4?
Stuck? Write out the sample space.
Call the four boys A, B, C, D.
List all unordered pairs: AB, AC, AD, BC, BD, CD
6 pairs.
Notice AB and BA were not counted separately — same pair.
(ii) Deriving the formula
Recall:
${}^{n}P_{r} = \dfrac{n!}{(n-r)!}$ (arranges in order)
For our example: ${}^{4}P_{2} = \dfrac{4!}{2!} = 12$. But that counts AB and BA separately.
Each pair of 2 can be arranged in $2! = 2$ ways. So divide out the $r!$:
${}^{4}C_{2} = \dfrac{{}^{4}P_{2}}{2!} = \dfrac{12}{2} = 6$ ✓
In general:
${}^{n}C_{r} = \dfrac{{}^{n}P_{r}}{r!} = \dfrac{n!}{r!\,(n-r)!}$
Must learn
1.${}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}$
2.Order doesn't matter — AB and BA count once.
YOU TRY · 1
Find ${}^{5}C_{2}$.
Work it out on paper first.
${}^{5}C_{2} = 10$
$10$
YOU TRY · 2
Find ${}^{7}C_{3}$.
Use the formula or your calculator's nCr button.
${}^{7}C_{3} = 35$
$35$
YOU TRY · 3
Find ${}^{10}C_{4}$.
Calculator OK from here on, but know how to do it by hand.
${}^{10}C_{4} = 210$
$210$
Section 2 of 11
Direct use — the Lotto question
How many ways can 6 numbers be chosen from 42?
This is the old Irish Lotto.
$n = 42, \quad r = 6$
${}^{42}C_{6} = \dfrac{42!}{6! \cdot 36!}$
$= 5{,}245{,}786$
Your calculator has an nCr button. Use it.
YOU TRY · 4
How many ways can 4 numbers be chosen from 12?
Identify $n$ and $r$, then formula or button.
$n = 12, \quad r = 4$
${}^{12}C_{4} = 495$
$495$
YOU TRY · 5
How many ways can 5 numbers be chosen from 20?
Calculator's fine.
${}^{20}C_{5} = 15{,}504$
$15{,}504$
YOU TRY · 6
How many 5-card hands can be dealt from a standard 52-card deck?
Order doesn't matter in a poker hand.
${}^{52}C_{5} = 2{,}598{,}960$
$2{,}598{,}960$
Section 3 of 11
When someone must be included
How many ways can 6 people be chosen from 10 if one particular person must get a place?
First instinct — wrong:
${}^{10}C_{6}$ (doesn't force anyone in)
${}^{10}C_{5}$ (wrong size — and still doesn't force anyone in)
The trick: lock that person in. They've taken a place — done. Now just pick the other 5 from the remaining 9.
${}^{9}C_{5} = 126$
Rule of thumb: when one is fixed in, subtract them from both $n$ AND $r$.
YOU TRY · 7
A team of 4 is picked from 8 players. The captain must play. How many teams are possible?
Lock the captain in. Now what's left?
Captain locked in. Pick 3 more from 7.
${}^{7}C_{3} = 35$
$35$
YOU TRY · 8
A committee of 5 is chosen from 12 people. Two specific members must be included. How many committees?
Two fixed in. Drop 2 from $n$ AND 2 from $r$.
Two locked in. Pick 3 more from 10.
${}^{10}C_{3} = 120$
$120$
YOU TRY · 9
A team of 3 is picked from 9. One particular player refuses to play. How many teams?
Different twist — this person is out, not in. Drop them from $n$ only.
That player out. Pick 3 from the remaining 8.
${}^{8}C_{3} = 56$
$56$
Section 4 of 11
Sub-division — AND and OR
Now we mix two groups. There are 5 men and 6 women who apply for 3 jobs. How many ways can the jobs be filled?
No restriction — anyone can take any of the 3 jobs.
First instinct — wrong:
$11 \times 10 \times 9 = 990$ (that's order — a permutation)
$6 \times 5 \times 4 \, + \, 5 \times 4 \times 3$ (guessing)
Straight answer — 5M + 6W = 11 people, pick 3:
${}^{11}C_{3} = 165$
(i) Exactly 2 men employed
Need: 2 men AND 1 woman.
AND means multiply.
${}^{5}C_{2} \times {}^{6}C_{1}$
$= 10 \times 6 = 60$
(ii) At least 2 men employed
"At least 2" means 2 OR 3 (can't pick more than 3 total).
OR means add.
Case A — 2M and 1W: ${}^{5}C_{2} \times {}^{6}C_{1} = 60$
Case B — 3M and 0W: ${}^{5}C_{3} \times {}^{6}C_{0} = 10 \times 1 = 10$
Total $= 60 + 10 = 70$
Must learn
ANDmultiply ($\times$)
ORadd ($+$)
YOU TRY · 10
4 boys and 5 girls. Pick 3. How many groups have exactly 2 boys?
2B AND 1G — multiply.
2B AND 1G: ${}^{4}C_{2} \times {}^{5}C_{1} = 30$
$30$
YOU TRY · 11
4 boys and 5 girls. Pick 3. How many groups have at least 1 boy?
Three cases — 1B+2G, 2B+1G, 3B+0G. OR them. Or use Total − No-boys.
Case A — 1B, 2G: ${}^{4}C_{1} \times {}^{5}C_{2} = 40$
Case B — 2B, 1G: ${}^{4}C_{2} \times {}^{5}C_{1} = 30$
Case C — 3B, 0G: ${}^{4}C_{3} \times {}^{5}C_{0} = 4$
Total $= 74$
$74$
YOU TRY · 12
7 men and 5 women apply for 4 posts. How many panels with exactly 3 men?
3M AND 1W — multiply.
3M AND 1W: ${}^{7}C_{3} \times {}^{5}C_{1} = 175$
$175$
Section 5 of 11
A useful shortcut
Compute ${}^{6}C_{0}$ and ${}^{6}C_{6}$.
${}^{6}C_{0} = \dfrac{6!}{0! \cdot 6!} = 1$
${}^{6}C_{6} = \dfrac{6!}{6! \cdot 0!} = 1$
Recall $0! = 1$.
In general:
${}^{n}C_{0} = {}^{n}C_{n} = 1$
Now ${}^{9}C_{2}$ and ${}^{9}C_{7}$:
${}^{9}C_{2} = \dfrac{9!}{2! \cdot 7!} = 36$
${}^{9}C_{7} = \dfrac{9!}{7! \cdot 2!} = 36$
Same answer! Why?
Choosing 2 to keep is the same as choosing 7 to throw away.
Must learn
1.${}^{n}C_{r} = {}^{n}C_{n-r}$
2.When $r$ is big, flip it. Saves work.
YOU TRY · 13
Find ${}^{100}C_{97}$ without a calculator.
97 is huge. Flip it.
${}^{100}C_{97} = {}^{100}C_{3}$ (flip)
$= 161{,}700$
$161{,}700$
YOU TRY · 14
Find ${}^{20}C_{18}$.
Flip first.
${}^{20}C_{18} = {}^{20}C_{2}$ (flip)
$= 190$
$190$
YOU TRY · 15
Find ${}^{12}C_{10}$.
Flip it.
${}^{12}C_{10} = {}^{12}C_{2}$ (flip)
$= 66$
$66$
Section 6 of 11
Tests with sections — at least from each
A test has 6 questions in Section A and 8 in Section B. You must do 7 questions, with at least 3 from each section. How many ways can the 7 be chosen?
"At least 3 from each" with 7 to pick, so the split must be:
3A + 4B or 4A + 3B
Why not 5A + 2B? Because 2 < 3 — fails the rule.
Same for 2A + 5B.
Two cases joined by OR (= +):
Case A — 3A + 4B: ${}^{6}C_{3} \times {}^{8}C_{4} = 20 \times 70 = 1400$
Case B — 4A + 3B: ${}^{6}C_{4} \times {}^{8}C_{3} = 15 \times 56 = 840$
Total $= 1400 + 840 = 2240$
YOU TRY · 16
A test has 5 questions in Section A and 7 in Section B. Do 6 questions, at least 2 from each section. How many ways?
List the splits that obey "at least 2 from each" while totalling 6.
Splits: 2A+4B, 3A+3B, 4A+2B.
2A+4B: ${}^{5}C_{2} \times {}^{7}C_{4} = 350$
3A+3B: ${}^{5}C_{3} \times {}^{7}C_{3} = 350$
4A+2B: ${}^{5}C_{4} \times {}^{7}C_{2} = 105$
Total $= 805$
$805$
YOU TRY · 17
4 algebra questions, 6 geometry questions. Do 5 in total with at least 2 of each. How many ways?
Splits totalling 5 with each $\geq 2$.
Splits: 2A+3G, 3A+2G.
2A+3G: ${}^{4}C_{2} \times {}^{6}C_{3} = 120$
3A+2G: ${}^{4}C_{3} \times {}^{6}C_{2} = 60$
Total $= 180$
$180$
Section 7 of 11
Equal numbers — and "more of one"
6 students and 9 teachers apply for 4 posts.
(i) Equal numbers of each
4 posts split evenly — 2 students AND 2 teachers.
${}^{6}C_{2} \times {}^{9}C_{2}$
$= 15 \times 36 = 540$
(ii) More students than teachers
With 4 posts, "more students" means 3S+1T or 4S+0T.
2S+2T is equal, not "more" — excluded.
Case A — 3S + 1T: ${}^{6}C_{3} \times {}^{9}C_{1} = 20 \times 9 = 180$
Case B — 4S + 0T: ${}^{6}C_{4} \times {}^{9}C_{0} = 15 \times 1 = 15$
Total $= 180 + 15 = 195$
YOU TRY · 18
5 boys and 7 girls. A team of 4 is selected with equal numbers of each. How many teams?
2 boys AND 2 girls.
2B AND 2G: ${}^{5}C_{2} \times {}^{7}C_{2} = 210$
$210$
YOU TRY · 19
Same 5 boys and 7 girls. Team of 4 with more boys than girls. How many teams?
3B+1G or 4B+0G.
Case A — 3B + 1G: ${}^{5}C_{3} \times {}^{7}C_{1} = 70$
Case B — 4B + 0G: ${}^{5}C_{4} \times {}^{7}C_{0} = 5$
Total $= 75$
$75$
YOU TRY · 20
4 senior and 8 junior players. A team of 5 has more seniors than juniors. How many teams?
Only 4 seniors exist — careful with 5S+0J.
Cases (sen > jun, total 5): 3S+2J, 4S+1J. No 5S+0J — only 4 seniors.
Case A — 3S + 2J: ${}^{4}C_{3} \times {}^{8}C_{2} = 112$
Case B — 4S + 1J: ${}^{4}C_{4} \times {}^{8}C_{1} = 8$
Total $= 120$
$120$
Section 8 of 11
Thick question — "won't play if"
A panel has 12 players including Mary and Jane. A team of 5 is selected.
Mary will not play if Jane plays. How many teams are possible?
Translate: Mary and Jane never both appear together.
(i) Direct method — case split
Three valid cases — Mary alone, Jane alone, or neither:
Case 1 — Mary plays, Jane doesn't: lock M in, pick 4 from the other 10 (excluding J).
${}^{10}C_{4} = 210$
Case 2 — Jane plays, Mary doesn't: lock J in, pick 4 from the other 10 (excluding M).
${}^{10}C_{4} = 210$
Case 3 — Neither plays: pick all 5 from the other 10.
${}^{10}C_{5} = 252$
Total $= 210 + 210 + 252 = 672$
(ii) Alternative — Total minus Both
Easier route. Take ALL possible teams, then subtract the banned ones (where both Mary AND Jane play together).
Total teams: ${}^{12}C_{5} = 792$
Both M and J play: lock both in, pick 3 from the other 10.
Banned: ${}^{10}C_{3} = 120$
$792 - 120 = 672$ ✓
When a question has an exclusion ("won't play if", "must not both"), Total − Banned usually saves work.
YOU TRY · 21
10 swimmers including Tom and Anne. A team of 4 is picked. Tom won't swim if Anne does. How many teams?
Try the Total − Both method first.
Total: ${}^{10}C_{4} = 210$
Both T and A play: pick 2 more from 8 → ${}^{8}C_{2} = 28$
$210 - 28 = 182$
$182$
YOU TRY · 22
15 students including Seán and Mick. A debate team of 6 is picked. Seán refuses if Mick is on the team. How many teams?
Total − Both. Watch the numbers — they get big.
Total: ${}^{15}C_{6} = 5005$
Both S and M play: pick 4 from other 13 → ${}^{13}C_{4} = 715$
$5005 - 715 = 4290$
$4290$
Section 9 of 11
Diagrams I — points on a circle
5 points lie on a circle, no 3 of them collinear.
Collinear = on the same straight line.
How many triangles can be formed?
A triangle needs 3 non-collinear points. Since no 3 here are collinear, any 3 of the 5 work.
Choose 3 from 5: ${}^{5}C_{3}$
$= 10$
The 10 triangles:
ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE
(i) How many of those triangles use A?
Lock A in. Pick the other 2 vertices from the remaining 4.
${}^{4}C_{2} = 6$
Sanity check — the list above: ABC, ABD, ABE, ACD, ACE, ADE. Six. ✓
YOU TRY · 23
7 points on a circle, no 3 collinear. How many triangles can be formed?
Any 3 of the 7 will do.
${}^{7}C_{3} = 35$
$35$
YOU TRY · 24
8 points on a circle, no 3 collinear. How many quadrilaterals can be formed?
A quadrilateral needs 4 vertices.
${}^{8}C_{4} = 70$
$70$
YOU TRY · 25
6 points on a circle, no 3 collinear. How many triangles use a particular point X?
Lock X in. Choose 2 more from the others.
X locked in. Choose 2 more from 5.
${}^{5}C_{2} = 10$ triangles use X.
$10$
Section 10 of 11
Diagrams II — triangles between parallel lines
3 points $A, B, C$ lie on a line $p$. 4 points $w, x, y, z$ lie on a parallel line $q$.
How many triangles can be formed using these 7 points?
First instinct — wrong:
${}^{7}C_{3} = 35$ (this counts triples that are all on one line — not triangles!)
Three points on the same straight line make NO triangle. So we must split into cases where the 3 points are NOT all on the same line:
2 points on $p$ AND 1 on $q$ or 1 on $p$ AND 2 on $q$
Case A — 2p, 1q: ${}^{3}C_{2} \times {}^{4}C_{1} = 3 \times 4 = 12$
Case B — 1p, 2q: ${}^{3}C_{1} \times {}^{4}C_{2} = 3 \times 6 = 18$
Total $= 12 + 18 = 30$
Quick check: $35 - 1 - 4 = 30$. (35 total triples, minus 1 from "all 3 on p" = ${}^{3}C_{3}$, minus 4 from "all 3 on q" = ${}^{4}C_{3}$.) ✓
YOU TRY · 26
4 points on line $L_1$, 5 points on parallel line $L_2$. How many triangles can be formed?
Split: 2 from one line AND 1 from the other.
Case A — 2 on $L_1$, 1 on $L_2$: ${}^{4}C_{2} \times {}^{5}C_{1} = 30$
Case B — 1 on $L_1$, 2 on $L_2$: ${}^{4}C_{1} \times {}^{5}C_{2} = 40$
Total $= 70$
$70$
YOU TRY · 27
5 points on one line, 6 points on a parallel line. How many triangles?
Two cases.
2 from first AND 1 from second: ${}^{5}C_{2} \times {}^{6}C_{1} = 60$
1 from first AND 2 from second: ${}^{5}C_{1} \times {}^{6}C_{2} = 75$
Total $= 135$
$135$
Section 11 of 11
Diagrams III — parallelograms
Same setup as §10 — 3 points $A, B, C$ on line $p$ and 4 points $w, x, y, z$ on parallel line $q$.
How many parallelograms can be formed?
A parallelogram needs 2 vertices on each line — one pair on $p$, the other pair on $q$.
First instinct:
2 from $p$ AND 2 from $q$:
${}^{3}C_{2} \times {}^{4}C_{2} = 3 \times 6 = 18$
But wait — that counts every "quadrilateral with 2 on each line", not just parallelograms.
For a true parallelogram, the gap on $p$ must equal the gap on $q$.
Carl's notes show this only works for matching equidistant pairs — count them carefully.
Suppose points are evenly spaced on each line. Let gaps on $p$ be $1$ unit (between consecutive points), so on $p$ the possible gap-sizes between two chosen points are $1$ (AB, BC) or $2$ (AC). On $q$ the gap-sizes are $1$ (wx, xy, yz), $2$ (wy, xz) or $3$ (wz).
For a parallelogram, gap on $p$ = gap on $q$:
Gap 1 on $p$: 2 pairs (AB, BC). Gap 1 on $q$: 3 pairs (wx, xy, yz). → $2 \times 3 = 6$
Gap 2 on $p$: 1 pair (AC). Gap 2 on $q$: 2 pairs (wy, xz). → $1 \times 2 = 2$
Gap 3 on $p$: 0 pairs. Gap 3 on $q$: 1 pair. → $0 \times 1 = 0$
Total $= 6 + 2 + 0 = 8$ parallelograms
Same gap-size on both lines is what makes it a parallelogram. If the gaps don't match, you just get a trapezium.
Diagram questions — remember
1.Triangles: 3 non-collinear points. Watch for points on a line.
2.Quadrilaterals between parallel lines: 2 from each line.
3.Parallelograms: equal gaps on each line.
SUM
The lot in one box
Combinations toolkit
1.${}^{n}C_{r} = \dfrac{n!}{r!\,(n-r)!}$ (order doesn't matter)
2.${}^{n}C_{r} = {}^{n}C_{n-r}$ (flip when $r$ is big)
3.${}^{n}C_{0} = {}^{n}C_{n} = 1$
4.AND $\to \times$, OR $\to +$
5."Must include $k$": drop $k$ from $n$ and from $r$.
6."Won't both": Total $-$ Both.
7."At least $k$": case split, then OR (add).
End of lesson
Combinations — HL · Mathslive.ie