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Difficulty: HardCombinations

A student library committee is selecting 55 distinct books from a shelf containing 77 novel titles and 55 biography titles. If 22 specific novel titles are mutually exclusive (they cannot both be selected together in the same combination), in how many ways can the selection of 55 books be made such that at least 33 novel titles are included?

  1. A
    426
  2. 436Answer
  3. C
    516
  4. D
    546

Answer

436
To find the number of valid ways, we first find the total number of ways to select at least 33 novels from 77 novels and 55 biographies, which equals 546546. Next, we determine how many of these combinations contain both of the restricted novels while still having at least 33 novels overall. There are 110110 such invalid selections. Subtracting 110110 from 546546 gives 436436 valid ways.

Step-by-Step Solution

1
Calculate the total combinations with at least 3 novels without any restriction.
Case 1 (33 Novels, 22 Biographies): (73)×(52)=35×10=350\binom{7}{3} \times \binom{5}{2} = 35 \times 10 = 350.
Case 2 (44 Novels, 11 Biography): (74)×(51)=35×5=175\binom{7}{4} \times \binom{5}{1} = 35 \times 5 = 175.
Case 3 (55 Novels, 00 Biographies): (75)×(50)=21×1=21\binom{7}{5} \times \binom{5}{0} = 21 \times 1 = 21.
Total unrestricted combinations = 350+175+21=546350 + 175 + 21 = 546.
Establishing the total pool of choices that satisfy the constraint of selecting at least 3 novels.
2
Calculate the invalid combinations where both restricted novels are selected together AND at least 3 novels are included.
If both restricted novels are selected, we have already chosen 22 novels. To reach a total of 55 books with at least 33 novels, we must pick 33 additional books from the remaining 1010 books (55 remaining novels and 55 biographies), EXCLUDING the scenario where all 33 additional books are biographies (which would leave us with only 22 novels in total).
Combinations of 33 books from 1010 remaining books = (103)=120\binom{10}{3} = 120.
Combinations of 33 biographies from 55 biographies = (53)=10\binom{5}{3} = 10.
Invalid combinations = 120−10=110120 - 10 = 110.
To apply the mutual exclusion constraint correctly, we must subtract only those invalid selections that also satisfy the condition of having at least 3 novels.
3
Subtract invalid combinations from total unrestricted combinations.
Valid combinations = 546−110=436546 - 110 = 436.
Subtracting the forbidden overlapping outcomes yields the net valid choices.

Key Concept

Combinations with multiple constraints and mutual exclusion
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Topics

Combinations

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