Permutation & Combination — Distribution Problems
Permutation & Combination
50
JEE Qs
18%
Hard
75
min
Always meticulously categorize the problem by asking: 'Are the items distinct or identical?' and 'Are the containers/recipients distinct or identical?' before attempting a solution.
🧮 Key Formulas
✅ Key Points for JEE
- 1The most crucial step is to correctly identify whether the items being distributed are distinct or identical, and whether the containers/recipients (boxes) are distinct or identical. This dictates the appropriate formula or method.
- 2For distinct items into distinct boxes, think about each item's independent choice. For distinct items into identical boxes, usually, consider partitions of items into groups and then assign them.
- 3For identical items into distinct boxes, use the 'Stars and Bars' method (also known as multinomial coefficient approach for non-negative integer solutions to x1 + ... + xr = n).
- 4For identical items into identical boxes, this is equivalent to finding partitions of an integer 'n' into 'r' parts. This is often solved by listing cases or using generating functions/recursion for advanced problems, as there's no simple direct formula usually taught at JEE level.
- 5Problems with conditions like 'at least one item in each box' or 'maximum capacity' often require applying the inclusion-exclusion principle or making initial assignments before using standard formulas.
⚠️ Common Mistakes
- ✕Confusing distinct items with identical items, or distinct boxes with identical boxes, leading to the application of incorrect formulas.
- ✕Incorrectly applying the 'Stars and Bars' formula (C(n+r-1, r-1)) to situations involving distinct items.
- ✕Failing to account for 'empty boxes allowed' vs. 'no empty boxes' conditions, especially when using inclusion-exclusion.
- ✕Overcounting or undercounting when dealing with identical boxes, as order of distribution among identical boxes does not matter.
- ✕Not properly breaking down complex problems into sub-cases when direct formulas are not applicable (e.g., when specific minimum/maximum conditions apply to individual boxes).
📝 Practice Questions
See allQ2. In a group of 3 girls and 4 boys, there are two boys B1 and B2 . The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B1 and B2 are not adjacent to each other, is : (1) 96 (2) 144 (3) 120 (4) 72
Q18.Let the shortest distance from (a, 0), a > 0, to the parabola y2 = 4x be 4 . Then the equation of the circle passing through the point (a, 0) and the focus of the parabola, and having its centre on the axis of the parabola is : (1) x2 + y2 −10x + 9 = 0 (2) x2 + y2 −6x + 5 = 0 (3) x2 + y2 −4x + 3 = 0 (4) x2 + y2 −8x + 7 = 0
Q25.The number of 3 -digit numbers, that are divisible by 2 and 3 , but not divisible by 4 and 9 , is______.
Q10.From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is ' M ', is : 2025 (22 Jan Shift 1) JEE Main Previous Year Paper (1) 5148 (2) 6084 (3) 4356 (4) 14950
Q13.The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is (1) 36000 (2) 37000 (3) 34000 (4) 35000
Q23.The number of ways, 5 boys and 4 girls can sit in a row so that either all the boys sit together or no two boys sit together, is -
NCERT Chapters
- Class 11 Maths Ch 7: Permutations and Combinations