IIT JEE Main Maths -Unit 4- Combinations (nCr)- Study Notes-New Syllabus
IIT JEE Main Maths -Unit 4- Combinations (nCr) – Study Notes – New syllabus
IIT JEE Main Maths -Unit 4- Combinations (nCr) – Study Notes -IIT JEE Main Maths – per latest Syllabus.
Key Concepts:
- Combinations — Definition, Formulae, and Properties (Including Selection with Restrictions)
- Permutations and Combinations — Mixed Problems (Selection + Arrangement)
Combinations — Definition, Formulae, and Properties (Including Selection with Restrictions)
A combination is a selection of some or all objects from a given set, without considering their order.
Example: Choosing 3 players from a team of 10 — the order in which they are chosen does not matter.
Formula for Combinations
If there are \( n \) distinct objects and we have to choose \( r \) of them, then the number of possible combinations is:
$ C(n, r) = nC_r = \dfrac{n!}{r!(n – r)!} $
Note: Since order does not matter, combinations are always fewer than permutations.
Relation between permutation and combination:
$ nP_r = nC_r \times r! $
Important Values
- \( nC_0 = nC_n = 1 \)
- \( nC_1 = nC_{n – 1} = n \)
Properties of Combinations
| Property | Formula |
|---|---|
| Symmetry Property | \( nC_r = nC_{n – r} \) |
| Pascal’s Identity | \( nC_r + nC_{r – 1} = (n + 1)C_r \) |
| Summation Property | \( \sum_{r=0}^n nC_r = 2^n \) |
Combinations with Restrictions
- If some particular object must always be included, reduce both \( n \) and \( r \) by 1:
\( \text{New combinations} = (n – 1)C_{r – 1} \) - If some particular object must always be excluded, reduce only \( n \):
\( \text{New combinations} = (n – 1)C_r \)
Combination of All Subsets
Total number of subsets of a set of \( n \) elements = \( 2^n \).
Proof: $\sum_{r=0}^{n} nC_r = 2^n$
Relationship Between Permutation and Combination
$ nP_r = nC_r \times r! $
This relation helps to convert problems involving arrangements into selection-based forms.
Key Takeaways for JEE:
- Combinations deal with selection, not order.
- Use \( nC_r = \dfrac{n!}{r!(n – r)!} \) as the standard formula.
- Apply symmetry: \( nC_r = nC_{n – r} \).
- For “at least” or “at most” type problems, sum over required values of \( r \).
- Combination questions often combine with restrictions or grouping logic.
Example
In how many ways can a committee of 3 members be formed from 6 people?
▶️ Answer / Explanation
We are selecting 3 people out of 6 → order doesn’t matter.
\( 6C_3 = \dfrac{6!}{3!3!} = \dfrac{720}{36} = 20 \)
Answer: 20 possible committees.
Example
From 8 men and 5 women, a committee of 4 persons is to be formed such that exactly 2 are women. In how many ways can this be done?
▶️ Answer / Explanation
Choose 2 women from 5 and 2 men from 8.
\( \text{Total ways} = 5C_2 \times 8C_2 = 10 \times 28 = 280 \)
Answer: 280 ways to form the committee.
Example
Out of 10 people, 4 are teachers. A group of 5 persons is to be formed such that at least 1 teacher is included. Find the number of ways.
▶️ Answer / Explanation
Total people = 10 (4 teachers + 6 non-teachers).
Step 1: Total ways to form any group of 5 people = \( 10C_5 = 252 \).
Step 2: Subtract cases with no teacher.
No teacher ⇒ choose all from 6 non-teachers ⇒ \( 6C_5 = 6 \).
Step 3: Required = \( 252 – 6 = 246 \)
Answer: 246 possible groups.
Permutations and Combinations — Mixed Problems (Selection + Arrangement)
Many real-world and JEE problems involve both selection (combinations) and arrangement (permutations).
In such problems, we first choose the required elements (using combinations) and then arrange them (using permutations).
$ \text{Total number of ways} = (\text{Ways of selection}) \times (\text{Ways of arrangement}) $
General Rule
If you must select \( r \) items out of \( n \) and then arrange the selected items, the total number of ways is:
$ nC_r \times r! = nP_r $
This is the direct link between combinations and permutations.
| Permutations vs. Combinations | ||
| Feature | Permutations | Combinations |
| Definition | Arrangements where order matters. | Selections where order doesn’t matter. |
| Formula | \( P(n,k)=\dfrac{n!}{(n-k)!} \) | \( C(n,k)=\dfrac{n!}{k!(n-k)!} \) |
| Examples | • Arranging Books on a Shelf • Assigning Specific Roles or Positions • Race Outcomes • Creating Passwords or Codes | • Choosing Friends for a Group Activity • Selecting Toppings for a Pizza • Drawing Cards in a Card Game • Forming a Committee |
Common JEE Scenarios
- Forming words or numbers — choose letters/digits first, then arrange.
- Committee formation with roles — select members, then assign positions.
- Seating arrangements — select who sits, then arrange them in line or circle.
Example
From 6 people, in how many ways can a group of 3 be selected and arranged in a row?
▶️ Answer / Explanation
Step 1: Select 3 people out of 6 → \( 6C_3 \) ways.
Step 2: Arrange 3 selected people in a row → \( 3! \) ways.
Total = \( 6C_3 \times 3! = 20 \times 6 = 120 \)
Answer: 120 ways.
Example
In how many ways can a committee of 4 be formed from 8 people such that one of them is appointed as a leader?
▶️ Answer / Explanation
Step 1: Select 4 people from 8 → \( 8C_4 \) ways.
Step 2: Appoint 1 of the 4 as leader → \( 4C_1 = 4 \) ways.
Total = \( 8C_4 \times 4 = 70 \times 4 = 280 \)
Answer: 280 different committees can be formed.
Example
From 8 friends, 5 are to be selected and seated around a circular table. In how many ways can this be done?
▶️ Answer / Explanation
Step 1: Select 5 friends out of 8 → \( 8C_5 = 56 \) ways.
Step 2: Arrange 5 people around a circular table → \( (5 – 1)! = 24 \) ways.
Step 3: Total = \( 56 \times 24 = 1344 \)
Answer: 1344 possible arrangements.
Special Cases & Tricks
- At least / At most cases → use summation across possible \( r \) values.
- Different roles (President, Secretary, etc.) → multiply by \( r! \) after selection.
- Identical objects → divide by factorial of repetitions.
- Seating around circle → use \( (n – 1)! \).
Example
From 10 students, in how many ways can a President, Vice-President, and Secretary be chosen?
▶️ Answer / Explanation
Step 1: Select 3 students from 10 → \( 10C_3 \).
Step 2: Assign 3 roles (arrangements) → \( 3! \).
Total = \( 10C_3 \times 3! = 120 \times 6 = 720 \)
Answer: 720 ways to assign positions.