| --- |
| source: ncert |
| topic: combinatorics |
| class_level: class_11 |
| chapter: ch7 |
| difficulty: intermediate |
| --- |
| |
| Class 11 | Ch7: Permutations and Combinations |
| FUNDAMENTAL COUNTING PRINCIPLE: |
| If task A can be done in m ways and task B in n ways, both together: m×n ways. |
| FACTORIAL: n!=n×(n-1)×(n-2)×...×2×1. 0!=1. 1!=1. |
| PERMUTATIONS (ordered arrangements): |
| nPr = n!/(n-r)! = n×(n-1)×...×(n-r+1). (r items from n, order matters.) |
| Permutations of n items = n! |
| Permutations with repetition: n^r (r items from n with repetition). |
| Permutations with identical items: n!/p!q!r!... (p identical of one type, q of another etc.) |
| CIRCULAR PERMUTATIONS: (n-1)! for n distinct items in circle. |
| COMBINATIONS (unordered selections): |
| nCr = n!/(r!(n-r)!) = nPr/r!. (r items from n, order does not matter.) |
| nCr = nC(n-r). nC0=nCn=1. nC1=n. |
| PASCAL'S IDENTITY: nCr + nCr-1 = (n+1)Cr. |
| SOLVED EXAMPLES: |
| Example 1: 5 boys and 3 girls sit in row, girls always together. |
| Treat 3 girls as 1 unit. Total units=6. Arrange 6 units: 6! ways. Girls arrange among themselves: 3! ways. |
| Total=6!×3!=720×6=4320. |
| Example 2: Find 7C3. |
| 7C3=7!/(3!×4!)=7×6×5/(3×2×1)=210. |
| Example 3: Prove nCr+nCr-1=(n+1)Cr. |
| nCr+nCr-1=n!/(r!(n-r)!)+n!/((r-1)!(n-r+1)!) |
| =n!/((r-1)!(n-r)!) × [1/r+1/(n-r+1)] |
| =n!/((r-1)!(n-r)!) × (n+1)/(r(n-r+1)) |
| =(n+1)!/(r!(n-r+1)!)=(n+1)Cr. Proved. |
| Example 4: How many ways to choose 3 from 7 if 2 specific must be included? |
| 2 fixed. Choose 1 from remaining 5: 5C1=5 ways. |
| COMMON MISTAKES: |
| Permutation: ORDER matters (abc≠bac). Combination: order does NOT matter. |
| Circular: fix one position, arrange remaining (n-1)!. |
| nCr = nC(n-r): 10C3=10C7=120. |
