📘 1.2 Permutations and combinations – arrange or choose?
Hello students! Today we’re learning two powerful counting tools: permutations (order matters) and combinations (order doesn’t matter). These are super common in exams – from arranging students in a line to picking a committee. I’ll explain every formula step by step and give you lots of practice. 🇪🇹 Let’s go!
👩🏫 Remember: the key question is “does order matter?” If you’re lining people up → permutation. If you’re just choosing a team → combination. We’ll see many examples.
🔁 Permutation – ordered arrangement
📌 A permutation of \(r\) objects from \(n\) distinct objects is an ordered list (sequence) of length \(r\). Notation: \(P(n,r)\) or \(_nP_r\).
\[ P(n,r) = n(n-1)(n-2)\cdots(n-r+1) = \frac{n!}{(n-r)!} \]
\[ P(n,r) = n(n-1)(n-2)\cdots(n-r+1) = \frac{n!}{(n-r)!} \]
For example, 3-permutations of \(\{A,B,C,D,E\}\) = \(P(5,3) = 5·4·3 = 60\). We listed them in the textbook. If all \(n\) objects are taken: \(P(n,n) = n!\) .
🏆 Example 1 (Exit‑exam style): 100 people enter a contest. How many ways to award first, second, third prize?
➡️ Order matters (gold, silver, bronze are different). \(P(100,3) = 100·99·98 = 970,200\).
(Think: 100 choices for 1st, 99 left for 2nd, 98 left for 3rd.)
➡️ Order matters (gold, silver, bronze are different). \(P(100,3) = 100·99·98 = 970,200\).
(Think: 100 choices for 1st, 99 left for 2nd, 98 left for 3rd.)
📸 Example 2: 5 students line up for a photo. How many orders?
That’s \(P(5,5)=5! = 120\).
That’s \(P(5,5)=5! = 120\).
╔══════╗ positions: [1st] [2nd] [3rd] [4th] [5th]
║ 5 ║ → 5 · 4 · 3 · 2 · 1 = 120
╚══════╝
🔁 Permutations with repetitions – different objects, unlimited supply
📌 The number of \(r\)-permutations from \(n\) distinct objects when repetition is allowed is simply \(n^r\).
🔐 Example 3: How many 4‑digit codes can be made using digits 0‑9 if digits can repeat?
\(10^4 = 10000\) (choose each digit independently).
\(10^4 = 10000\) (choose each digit independently).
✍️ Exam‑style MCQs – Permutations
1️⃣ There are 8 runners. Gold, silver, bronze medals are awarded. How many different outcomes? (no ties)
A) 336 B) 56 C) 512 D) 40320
Correct: A) 336
\(P(8,3) = 8·7·6 = 336\). Order of medals matters → permutation.
\(P(8,3) = 8·7·6 = 336\). Order of medals matters → permutation.
2️⃣ How many ways can 6 people sit in a row?
A) 720 B) 120 C) 36 D) 6
Correct: A) 720
\(P(6,6)=6! = 720\).
\(P(6,6)=6! = 720\).
3️⃣ Find \(n\) if \(P(n,2)=72\).
A) 8 B) 9 C) 10 D) 12
Correct: B) 9
\(P(n,2)=n(n-1)=72\) → \(n^2-n-72=0\) → \((n-9)(n+8)=0\) → \(n=9\).
\(P(n,2)=n(n-1)=72\) → \(n^2-n-72=0\) → \((n-9)(n+8)=0\) → \(n=9\).
🧩 Combinations – order doesn’t matter
📌 An \(r\)-combination is a subset of size \(r\). Number: \[ C(n,r) = \binom{n}{r} = \frac{n!}{r!(n-r)!} \]
Example: 3-combinations of \(\{a,b,c,d,e\}\): \(\binom{5}{3}=10\). The list is in the textbook: {a,b,c}, {a,b,d}, … {c,d,e}.
🃏 Example 4 (classic): How many poker hands (5 cards) from a 52‑card deck?
\(C(52,5) = \frac{52!}{5!47!} = 2,598,960\).
\(C(52,5) = \frac{52!}{5!47!} = 2,598,960\).
👩🏫 Key relationship: \(C(n,r) = C(n, n-r)\). Example: choosing 3 winners from 10 = choosing 7 losers.
✍️ Exam‑style MCQs – Combinations
4️⃣ A farmer buys 3 cows from 6 cows, 2 pigs from 5 pigs, 4 hens from 8 hens. How many choices?
A) 12000 B) 14000 C) 2100 D) 4200
Correct: B) 14000
\(\binom{6}{3}\binom{5}{2}\binom{8}{4} = 20·10·70 = 14000\).
\(\binom{6}{3}\binom{5}{2}\binom{8}{4} = 20·10·70 = 14000\).
5️⃣ How many ways to select a 4‑member committee from 10 students?
A) 210 B) 5040 C) 151200 D) 420
Correct: A) 210
\(C(10,4) = \frac{10·9·8·7}{4·3·2·1} = 210\).
\(C(10,4) = \frac{10·9·8·7}{4·3·2·1} = 210\).
🔄 Circular permutations – sitting around a table
📌 Number of ways to arrange \(n\) distinct objects in a circle = \((n-1)!\)
(for \(r\) out of \(n\) in a circle: \(\frac{P(n,r)}{r}\))
(for \(r\) out of \(n\) in a circle: \(\frac{P(n,r)}{r}\))
🧑🤝🧑 Example 5: 5 men & 5 women sit around a table. (i) no restriction → \((10-1)! = 9! = 362880\).
(ii) no two women together: first seat 5 men (4! ways), then 5 women choose 5 gaps (5! ways) → \(4!·5! = 24·120 = 2880\).
(ii) no two women together: first seat 5 men (4! ways), then 5 women choose 5 gaps (5! ways) → \(4!·5! = 24·120 = 2880\).
6️⃣ How many ways to seat 6 people at a round table?
A) 120 B) 720 C) 600 D) 24
Correct: A) 120
\((6-1)! = 5! = 120\).
\((6-1)! = 5! = 120\).
🔤 Permutations with indistinguishable objects
📌 If there are \(n\) objects with \(n_1\) identical of type1, \(n_2\) identical of type2, … , total permutations = \(\frac{n!}{n_1!\,n_2!\cdots}\).
📖 Example 6 (word BENZENE): 7 letters, E appears 3 times, N appears 2 times → \(\frac{7!}{3!2!} = 420\).
📖 Example 7 (MISSISSIPPI): 11 letters: M:1, I:4, S:4, P:2 → \(\frac{11!}{1!4!4!2!} = 34650\).
7️⃣ How many distinct arrangements of the word “BENZENE”?
A) 5040 B) 420 C) 840 D) 1260
Correct: B) 420
\(\frac{7!}{3!2!}=420\).
\(\frac{7!}{3!2!}=420\).
📐 Binomial theorem and binomial coefficients
\[(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k}y^k\]
🔹 Example 8: Expand \((2a-3b)^4\) = \(\binom{4}{0}(2a)^4(-3b)^0 + … + \binom{4}{4}(-3b)^4\) = \(16a^4 -96a^3b +216a^2b^2 -216ab^3 +81b^4\).
🔹 Example 9: Coefficient of \(x^5y^4\) in \((x+y)^9\) = \(\binom{9}{4} = 126\).
📝 Exercise 1.2 – selected answers
| Question | Answer / hint |
|---|---|
| 1. List all permutations of {a,b,c} | abc, acb, bac, bca, cab, cba (6 permutations) |
| 2. Permutations of {a,…,g} (7 letters) | 7! = 5040 |
| 3. Permutations ending with a | 6! = 720 (fix a at end, arrange the other 6) |
| 5a. C(5,1) | 5 |
| 5b. C(5,3) | 10 (same as C(5,2)=10) |
| 8a. 4‑member committee from 10 students | C(10,4)=210 |
| 8b. 2 men, 2 women (6 men,4 women) | C(6,2)·C(4,2)=15·6=90 |
| 9a. P(n,2)=110 → n | n(n-1)=110 → n=11 |
| 9c. P(n,4)=12P(n,2) → n | n(n-1)(n-2)(n-3)=12n(n-1) → cancel n(n-1) → (n-2)(n-3)=12 → n=6 |
| 10. Integers >5400, digits distinct, no 2 or 7 | Break into cases (5‑digit & 4‑digit) … final answer 240 |
| 16. Expand (x+y)^6 | x⁶+6x⁵y+15x⁴y²+20x³y³+15x²y⁴+6xy⁵+y⁶ |
📌 Quick reference – exam tips
- Permutation – order counts (passwords, races, line).
- Combination – order ignored (committee, handshake).
- Circular – fix one person to break symmetry.
- Identical objects – divide by factorial of counts.
- Pascal’s identity: \(\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}\).
8️⃣ (mixed) In how many ways can 5 men and 5 women sit around a table with no two women adjacent?
A) 2880 B) 14400 C) 86400 D) 1200
Correct: A) 2880
Seat 5 men: (5-1)! = 4! = 24. 5 women into 5 gaps: 5! = 120. Multiply: 24·120 = 2880.
Seat 5 men: (5-1)! = 4! = 24. 5 women into 5 gaps: 5! = 120. Multiply: 24·120 = 2880.
9️⃣ Find the number of 5‑letter words from letters a‑h with no repeated letters.
A) 6720 B) 32768 C) 336 D) 56
Correct: A) 6720
\(P(8,5) = 8·7·6·5·4 = 6720\).
\(P(8,5) = 8·7·6·5·4 = 6720\).
⚡ More binomial coefficients
\(\binom{20}{17} = \binom{20}{3} = \frac{20·19·18}{3·2·1}=1140\).
\(\sum_{k=0}^{n}\binom{n}{k}=2^n\). For \(n=4\) sum = \(1+4+6+4+1=16=2^4\).
\(\sum_{k=0}^{n}\binom{n}{k}=2^n\). For \(n=4\) sum = \(1+4+6+4+1=16=2^4\).
You’ve got this! Practice makes perfect. In the next section we’ll tackle the binomial theorem in depth. Stay sharp! 🇪🇹