Permutations & Combinations – WBJEE Mathematics Study Notes
Audience – WBJEE candidates (Class 12)
Format – Markdown, clear sections, key points, worked example, quick revision
1. Simple Explanation
| Concept | What it means | When to use |
|---|---|---|
| Permutation | An ordered arrangement of objects. | Counting ways to line up people, seat arrangements, etc. |
| Combination | An unordered selection of objects. | Choosing a team, picking lottery numbers, etc. |
Why it matters – Many WBJEE problems ask “how many ways can X be arranged/selected?” Understanding the difference between order (permutation) and order‑independent (combination) is the key.
2. Key Points & Formulas
| Category | Formula | Conditions | Example |
|---|---|---|---|
| Permutations of distinct objects | (P(n,r)=\dfrac{n!}{(n-r)!}) | (r\le n) | 5 books arranged in 3 places |
| Combinations of distinct objects | (C(n,r)=\binom{n}{r}=\dfrac{n!}{r!(n-r)!}) | (r\le n) | Choose 2 friends from 4 |
| Permutations with repetition | (P(n,r)=n^r) | Each spot independent | 3‑digit PIN using digits 0‑9 |
| Combinations with repetition (stars & bars) | (C(n+r-1,r)=\binom{n+r-1}{r}) | Objects identical, unlimited supply | Distribute 5 identical candies to 3 kids |
| Derangements | (!n = n!\sum_{k=0}^n \frac{(-1)^k}{k!}) | Arrange n items so none in original place | Secret Santa |
| Inclusion–Exclusion | (\displaystyle | A_1\cup\dots\cup A_n | =\sum |
| Permutations of multiset (identical objects) | (\displaystyle \frac{n!}{n_1!n_2!\dots n_k!}) | (n_i) identical items of type i | Arrange “AAB” |
| Combinations of multiset (identical objects) | (\displaystyle \binom{n}{r}) with restrictions | Choose r objects from multiset | 2 apples + 3 oranges, choose 3 fruits |
3. Worked Example
Problem (WBJEE style)
A committee of 4 people is to be formed from 5 men and 3 women. How many committees contain at least 2 women?
Step‑by‑Step
-
Total committees (any mix):
(\displaystyle \binom{8}{4}=70) -
Exclude committees with 0 or 1 woman
- 0 women → all men: (\binom{5}{4}=5)
- 1 woman → choose 1 of 3 women and 3 of 5 men: (\binom{3}{1}\binom{5}{3}=3\times10=30)
-
Desired count
[ 70 - (5 + 30) = 35 ]
Answer: 35 committees contain at least two women.
Quick Verification
-
Check with direct counting:
- 2 women: (\binom{3}{2}\binom{5}{2}=3\times10=30)
- 3 women: (\binom{3}{3}\binom{5}{1}=1\times5=5)
- 4 women: impossible (only 3 women)
Sum: (30+5=35). ✔️
4. Quick Revision Summary
| Topic | Main Formula | Typical Use |
|---|---|---|
| Permutations (distinct) | (P(n,r)=n!/(n-r)!) | Order matters |
| Combinations (distinct) | (C(n,r)=\binom{n}{r}) | Order doesn’t matter |
| Repetition allowed | (P(n,r)=n^r,\ C(n+r-1,r)=\binom{n+r-1}{r}) | Repeating digits, stars & bars |
| Derangements | (!n = n! \sum_{k=0}^n (-1)^k/k!) | “None in original place” |
| Multiset permutations | (\frac{n!}{n_1!n_2!\dots}) | Identical items |
| Inclusion–Exclusion | ( | A_1\cup\dots |
| Common pitfalls | 1. Forgetting order vs no order 2. Miscounting identical items 3. Assuming unlimited supply when not |
Quick Practice Questions (try before checking)
- Arrange the letters of the word “BANANA” so that the two N’s are not adjacent.
- A 5‑digit lock can use digits 0–9. How many codes are possible if the first digit is not 0?
- From 10 students, how many ways can a sub‑committee of 3 be chosen if at least one girl must be included? (Assume 4 girls, 6 boys.)
- How many ways to distribute 7 identical candies to 4 children, if each child gets at least one candy?
*(Answer after solving: 1. 60, 2. 9 000