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JEE Mathematics 2024 Question with Solution

The number of ways of arranging 88 identical books into 44 identical shelves where any number of shelves may remain empty is equal to:

  • A

    1818

  • B

    1616

  • C

    1212

  • D

    1515

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: 88 identical books and 44 identical shelves, with empty shelves allowed.

Find: The number of ways to arrange the books into the shelves.

Since the shelves are identical and the books are also identical, this is equivalent to counting the partitions of 88 into at most 44 non-negative parts.

From the solution, the valid partitions are:

  • 8=88=8
  • 8=7+18=7+1
  • 8=6+28=6+2
  • 8=6+1+18=6+1+1
  • 8=5+38=5+3
  • 8=5+2+18=5+2+1
  • 8=5+1+1+18=5+1+1+1
  • 8=4+48=4+4
  • 8=4+3+18=4+3+1
  • 8=4+2+28=4+2+2
  • 8=4+2+1+18=4+2+1+1
  • 8=3+3+28=3+3+2
  • 8=3+3+1+18=3+3+1+1
  • 8=3+2+2+18=3+2+2+1
  • 8=2+2+2+28=2+2+2+2

Counting by empty shelves

Given: 88 identical books and 44 identical shelves.

Find: Total number of distinct arrangements when any number of shelves may remain empty.

Count arrangements according to the number of empty shelves.

If 33 shelves are empty:

(8,0,0,0)  ⟹  1 way(8,0,0,0) \implies 1 \text{ way}

If 22 shelves are empty:

(7,1,0,0),(6,2,0,0),(5,3,0,0),(4,4,0,0)  ⟹  4 ways(7,1,0,0), (6,2,0,0), (5,3,0,0), (4,4,0,0) \implies 4 \text{ ways}

If 11 shelf is empty:

(6,1,1,0),(5,2,1,0),(4,3,1,0),(4,2,2,0),(3,3,2,0)  ⟹  5 ways(6,1,1,0), (5,2,1,0), (4,3,1,0), (4,2,2,0), (3,3,2,0) \implies 5 \text{ ways}

If 00 shelves are empty:

(5,1,1,1),(4,2,1,1),(3,3,1,1),(3,2,2,1),(2,2,2,2)  ⟹  5 ways(5,1,1,1), (4,2,1,1), (3,3,1,1), (3,2,2,1), (2,2,2,2) \implies 5 \text{ ways}

Therefore,

Total=1+4+5+5=15\text{Total} = 1+4+5+5 = 15

Hence, the correct option is D.

Common mistakes

  • Treating the shelves as distinct is incorrect because interchanging identical shelves does not create a new arrangement. Count partitions, not ordered distributions.

  • Ignoring empty shelves is wrong because the question explicitly allows shelves to remain empty. So arrangements with fewer than 44 non-zero parts must also be included.

  • Missing the case 2+2+2+22+2+2+2 leads to an undercount. When all four identical shelves are used, equal group sizes are also valid partitions.

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