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Difficulty: Very hardPermutations

How many distinct five-digit even numbers greater than 50,00050,000 can be formed using the digits 2,3,5,6,7,2, 3, 5, 6, 7, and 88 if no digit may be repeated in any number?

  1. 240Answer
  2. B
    288
  3. C
    480
  4. D
    40

Answer

The total number of distinct five-digit even numbers greater than 50,000 is 240.
The correct total is 240. Because the first digit requirement (greater than 50,000 requires 5, 6, 7, or 8) and the last digit requirement (even requires 2, 6, or 8) share common elements (6 and 8), the problem must be split into two mutually exclusive cases: when the first digit is odd (2 choices for d1d_1, 3 for d5d_5, and 4P3=24^4P_3 = 24 for the middle digits, giving 2×3×24=1442 \times 3 \times 24 = 144) and when the first digit is even (2 choices for d1d_1, 2 for d5d_5, and 4P3=24^4P_3 = 24 for the middle digits, giving 2×2×24=962 \times 2 \times 24 = 96). Adding these yields 144+96=240144 + 96 = 240.

Step-by-Step Solution

1
Identify constraints on the positions of a five-digit number d1d2d3d4d5d_1 d_2 d_3 d_4 d_5.
For the number to be greater than 50,000, d1∈{5,6,7,8}d_1 \in \{5, 6, 7, 8\}. For the number to be even, d5∈{2,6,8}d_5 \in \{2, 6, 8\}.
The first digit determines the magnitude and the last digit determines parity.
2
Analyze Case 1 where the first digit d1d_1 is odd (55 or 77).
There are 2 choices for d1d_1. All 3 even digits remain available for d5d_5 (3 choices). The remaining 3 middle positions (d2,d3,d4d_2, d_3, d_4) can be arranged from the 4 remaining unused digits in 4P3=24^4P_3 = 24 ways. Total for Case 1 = 2×3×24=1442 \times 3 \times 24 = 144.
Choosing an odd first digit does not reduce the pool of available even digits for the units position.
3
Analyze Case 2 where the first digit d1d_1 is even (66 or 88).
There are 2 choices for d1d_1. Since one even digit is used in d1d_1, 2 even digits remain for d5d_5 (2 choices). The 3 middle positions are filled from the 4 remaining digits in 4P3=24^4P_3 = 24 ways. Total for Case 2 = 2×2×24=962 \times 2 \times 24 = 96.
Selecting an even digit for the ten-thousands place directly reduces the available choices for the units place.
4
Sum the possibilities from both mutually exclusive cases.
Total valid numbers = 144+96=240144 + 96 = 240.
Applying the addition principle of counting over disjoint cases.

Key Concept

Conditional permutations with non-disjoint position restrictions
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Topics

Permutations

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