Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

If n is the number of ways five different employees can sit into four indistinguishable offices where any office may have any number of persons including zero, then n is equal to:

Solution & Explanation

### Related Formula The number of ways to distribute m distinct items into k indistinguishable groups is given by the sum of Stirling numbers of the second kind: sum_r=1^k S(m, r) ### Core Logic Since the offices are **indistinguishable**, we look at partitions of the number 5 into at most 4 parts. Let's analyze each case based on distribution patterns: ### Step 1: Calculate Case Combinations - **Case 1: 5, 0, 0, 0** All employees in one office: frac5!5! = 1 text way - **Case 2: 4, 1, 0, 0** Four employees in one office, one in another: frac5!4!1! = 5 text ways - **Case 3: 3, 2, 0, 0** Three employees in one office, two in another: frac5!3!2! = 10 text ways - **Case 4: 3, 1, 1, 0** Three in one, and one each in two other offices. Since the two single-person offices are identical: frac5!3!1!1! cdot 2! = 10 text ways - **Case 5: 2, 2, 1, 0** Two groups of two, and one group of one. Since the two two-person offices are identical: frac5!2!2!1! cdot 2! = 15 text ways - **Case 6: 2, 1, 1, 1** One group of two, three groups of one. Since the three single-person offices are identical: frac5!2!(1!)^3 cdot 3! = 10 text ways ### Step 2: Total Sum Summing all these distinct non-overlapping partition groups: textTotal ways n = 1 + 5 + 10 + 10 + 15 + 10 = 51 ### Pattern Recognition Sees: Distinct items placed into identical/indistinguishable containers. Trap: When dividing items into groups of equal sizes (like two groups of 2 or three groups of 1), you must divide by the factorial of the frequency of those groups (2! and 3!) to avoid overcounting permutations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Permutations and Combinations

Reference Study Guides

Permutations And Combinations

More Permutations and Combinations Previous-Year Questions — Page 6

Q17 jee_main_2024_31_jan_evening Combinations and Permutations Formula
If for some m, n; ^6C_m + 2(^6C_m+1) + ^6C_m+2 > ^8C_3 and ^n-1P_3 : ^nP_4 = 1:8, then ^nP_m+1 + ^n+1C_m is equal to
  • A. 380
  • B. 376
  • C. 384
  • D. 372

Solution

### Related Formula ^nC_r + ^nC_r-1 = ^n+1C_r ### Core Logic Simplify the binomial combination: ^6C_m + 2(^6C_m+1) + ^6C_m+2 = (^6C_m + ^6C_m+1) + (^6C_m+1 + ^6C_m+2) Using Pascal's rule, this becomes: ^7C_m+1 + ^7C_m+2 = ^8C_m+2 Given condition: ^8C_m+2 > ^8C_3 = 56. For N=8, the central combinations yield the maximum value: ^8C_4 = 70. Others like ^8C_5 = 56, which is not strictly greater than 56. So m + 2 = 4 implies m = 2. Solve the permutations ratio: frac^n-1P_3^nP_4 = frac18 frac(n-1)(n-2)(n-3)n(n-1)(n-2)(n-3) = frac18 implies frac1n = frac18 implies n = 8 Calculate the target expression: ^nP_m+1 + ^n+1C_m = ^8P_3 + ^9C_2 = (8 times 7 times 6) + frac9 times 82 = 336 + 36 = 372 ### Pattern Recognition Binomial coefficient reduction using Pascal's triangle quickly collapses expanded nCr sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations
Q23 jee_main_2024_31_jan_morning Word Formation
The total number of words (with or without meaning) that can be formed out of the letters of the word 'DISTRIBUTION' taken four at a time, is equal to
Numerical Answer. Answer: 3734 to 3734

Solution

### Core Logic Letters in 'DISTRIBUTION': I(3), T(2), D, S, R, B, U, O, N. There are 9 distinct letters. ### Step 1: Case Analysis **Case 1: 3 alike, 1 distinct** Selection: Choose the letter 'I' (^1C_1) and 1 from the remaining 8 distinct letters (^8C_1). Arrangement: ^8C_1 times frac4!3! = 8 times 4 = 32. **Case 2: 2 alike of one kind, 2 alike of another kind** Since only 'I' and 'T' appear at least twice, we must choose both. Arrangement: ^2C_2 times frac4!2!2! = 1 times 6 = 6. ### Step 2: Further Cases **Case 3: 2 alike, 2 distinct** Selection: Choose 1 from the 2 repeated sets (^2C_1) and 2 from the remaining 8 distinct letters (^8C_2). Arrangement: ^2C_1 times ^8C_2 times frac4!2! = 2 times 28 times 12 = 672. **Case 4: All 4 distinct** Selection: Choose 4 from the 9 distinct letters (^9C_4). Arrangement: ^9C_4 times 4! = 126 times 24 = 3024. ### Step 3: Total Words Total = 3024 + 672 + 6 + 32 = 3734. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Permutations and Combinations

More Permutations and Combinations Questions — jee_main_2024_01_february_morning

Practice all Permutations and Combinations previous-year questions →

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Under Development
PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.

Loading Questions... Browse Topics
Latest from the Blog
View all →

Loading articles...