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Zorluk: ZorPermutations, Combinations, and Fundamental Counting Principle

A tech company's quality assurance division needs to form a 5-member project panel selected from a pool of 6 software engineers and 4 hardware engineers. The panel must include at least 2 software engineers and at least 1 hardware engineer. However, 2 specific software engineers, Engineer XX and Engineer YY, refuse to serve on the same panel together. How many different valid 5-member panels can be formed?

  1. A
    136
  2. 188Cevap
  3. C
    196
  4. D
    240
  5. E
    252

Cevap

188 valid panels can be formed.
The total number of panels satisfying the software and hardware role distribution rules is 240. Among these, exactly 52 panels contain both Engineer X and Engineer Y. Subtracting 52 from 240 results in 188 valid panels.

Adım Adım Çözüm

1
Determine valid software and hardware engineer compositions for a 5-member panel.
Three compositions satisfy the requirement of at least 2 software engineers (S) and at least 1 hardware engineer (H): (4S, 1H), (3S, 2H), and (2S, 3H).
Panels must have 5 total members adhering to the specified minimum headcount limits.
2
Calculate the total number of panels satisfying composition requirements without considering the exclusion restriction.
For (4S, 1H): (64)×(41)=15×4=60\binom{6}{4} \times \binom{4}{1} = 15 \times 4 = 60. For (3S, 2H): (63)×(42)=20×6=120\binom{6}{3} \times \binom{4}{2} = 20 \times 6 = 120. For (2S, 3H): (62)×(43)=15×4=60\binom{6}{2} \times \binom{4}{3} = 15 \times 4 = 60. Total composition-valid panels = 60+120+60=24060 + 120 + 60 = 240.
Apply the combination formula (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!} and the Fundamental Counting Principle.
3
Calculate the number of invalid panels that contain both Engineer X and Engineer Y.
If Engineer X and Engineer Y are both included (2 S), 3 remaining panel members must be selected from the remaining 4 software engineers and 4 hardware engineers. For (4S, 1H): (42)×(41)=6×4=24\binom{4}{2} \times \binom{4}{1} = 6 \times 4 = 24. For (3S, 2H): (41)×(42)=4×6=24\binom{4}{1} \times \binom{4}{2} = 4 \times 6 = 24. For (2S, 3H): (40)×(43)=1×4=4\binom{4}{0} \times \binom{4}{3} = 1 \times 4 = 4. Total invalid panels = 24+24+4=5224 + 24 + 4 = 52.
Isolating combinations that contain both restricted engineers allows simple subtraction from the total.
4
Subtract invalid panels from the total composition-valid panels.
240−52=188240 - 52 = 188.
This leaves only the panels that satisfy both composition and exclusion rules.

Anahtar Kavram

Combinations with multi-group minimum constraints and pair exclusion
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Konular

Permutations, Combinations, and Fundamental Counting Principle

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