Grouping & Distribution in Permutations & Combinations
Master the critical distinction between identical vs distinct groups with essential formulas and JEE-level examples.
The Critical Distinction: Identical vs Distinct Groups
The most common mistake in P&C problems is confusing identical groups (like heaps, teams without names) with distinct groups (like labeled boxes, named teams). This distinction changes the counting method completely.
🎯 Key Insight:
Identical Groups: Arrangements among groups don't create new cases
Distinct Groups: Each group is unique, so arrangements matter
Dividing n distinct objects into r groups of given sizes
When groups are unlabeled/identical and group sizes are specified
Formula:
$$ \frac{n!}{p_1! \times p_2! \times \cdots \times p_k!} $$
where $p_1, p_2, \ldots, p_k$ are the sizes of each group
Example:
Divide 10 distinct books into 3 groups of sizes 4, 3, and 3
Step 1: Apply formula: $\frac{10!}{4! \times 3! \times 3!}$
Step 2: Since two groups have same size (3 and 3), divide by $2!$
Step 3: Final answer: $\frac{10!}{4! \times 3! \times 3! \times 2!} = 2100$
💡 Remember:
When groups have identical sizes, divide by the factorial of number of groups with that size to avoid overcounting.
Distributing n distinct objects into r distinct boxes
When each group/box is labeled/distinct (like Box A, Box B, Box C)
Formula:
$$ r^n $$
Each object has r choices, and choices are independent
Example:
Distribute 5 different balls into 3 distinct boxes
Step 1: Each ball has 3 choices (Box 1, Box 2, or Box 3)
Step 2: Total ways: $3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243$
Step 3: Empty boxes are allowed in this case
Variation - No Empty Boxes:
Distribute 5 different balls into 3 distinct boxes with no box empty
Step 1: Use inclusion-exclusion: $3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$
Step 2: Calculate: $243 - 3 \times 32 + 3 \times 1 = 243 - 96 + 3 = 150$
⚡ Quick Comparison: Identical vs Distinct Groups
| Situation | Identical Groups | Distinct Groups |
|---|---|---|
| Dividing 6 books into 2 groups of 3 each | $\frac{6!}{3!3!2!} = 10$ ways | $\frac{6!}{3!3!} = 20$ ways |
| Distributing 4 balls into 2 groups | Stirling numbers concept | $2^4 = 16$ ways |
| Forming 3 teams from 9 players | Divide by 3! if teams unlabeled | No division if teams have names |
Distributing n distinct objects into r identical boxes
When boxes are identical/unlabeled and empty boxes are allowed
Concept:
Stirling Numbers of the Second Kind
Denoted by $S(n, r)$ - number of ways to partition n distinct objects into r non-empty unlabeled subsets
Stirling Number Formula:
$$ S(n, r) = \frac{1}{r!} \sum_{k=0}^{r} (-1)^k \binom{r}{k} (r-k)^n $$
Example:
Distribute 4 distinct balls into 2 identical boxes (empty allowed)
Step 1: Cases: 1 box used or 2 boxes used
Step 2: 1 box used: All balls in one box = 1 way
Step 3: 2 boxes used: $S(4, 2) = \frac{1}{2!}[2^4 - 2 \cdot 1^4] = \frac{1}{2}[16-2] = 7$
Step 4: Total: $1 + 7 = 8$ ways
🎯 JEE Focus:
Understand the conceptual difference - you won't need to compute large Stirling numbers, but you should know when the concept applies.
🚀 Problem-Solving Framework
Ask These Questions:
- Are the groups labeled or identical?
- Are empty groups allowed?
- Are group sizes specified or arbitrary?
- Are objects distinct or identical?
Common Patterns:
- Teams with names = distinct groups
- Heaps/piles = identical groups
- Labeled boxes = distinct groups
- Unlabeled boxes = identical groups
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. In how many ways can 12 different books be divided equally among 3 students?
2. How many ways to distribute 7 different balls into 4 identical boxes with no box empty?
3. Divide 8 people into 4 groups of 2 each for a doubles tennis tournament.
Mastered These Concepts?
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