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Number of ways of arranging 8 identical books into 4 identical shelves where any number of shelves may remain empty is equal to

Solution & Explanation

### Related Formula Since both the objects (books) and the containers (shelves) are identical, this problem is equivalent to finding the number of partitions of the integer 8 into at most 4 parts. ### Core Logic Let us systematically list out all the possible distributions based on the number of empty shelves: 1. **3 Shelves empty:** * (8, 0, 0, 0) rightarrow 1text way 2. **2 Shelves empty:** * (7, 1, 0, 0) * (6, 2, 0, 0) * (5, 3, 0, 0) * (4, 4, 0, 0) rightarrow 4text ways 3. **1 Shelf empty:** * (6, 1, 1, 0) * (5, 2, 1, 0) * (4, 3, 1, 0) * (4, 2, 2, 0) * (3, 3, 2, 0) rightarrow 5text ways 4. **0 Shelves empty:** * (5, 1, 1, 1) * (4, 2, 1, 1) * (3, 3, 1, 1) * (3, 2, 2, 1) * (2, 2, 2, 2) rightarrow 5text ways ### Step 1: Total Computations Summing all these cases together: textTotal ways = 1 + 4 + 5 + 5 = 15text ways ### Pattern Recognition Be very careful to identify if containers/objects are identical or distinct. Identical into identical means simple partition of integers. Listing them in descending order ensures no partition is missed or duplicated. ### 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

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