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Zorluk: ZorCombinations

A panel of 55 members is to be selected from 1010 eligible candidates. If 22 specific candidates refuse to serve on the panel together, in how many different ways can the panel be formed?

Cevap: 196

Cevap

The panel can be formed in 196 different ways.
To find the number of valid panels, use complementary counting. First, compute the total number of ways to pick any 5 candidates from 10 without restrictions: (105)=252\binom{10}{5} = 252. Next, find the number of invalid panels that contain both restricted candidates; since 2 candidates are already placed, pick the remaining 3 members from the remaining 8 candidates: (83)=56\binom{8}{3} = 56. Subtracting these invalid panels from the total gives 252−56=196252 - 56 = 196.

Adım Adım Çözüm

1
Calculate the total possible combinations without any restrictions.
(105)=10×9×8×7×65×4×3×2×1=252\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252
The combinations formula (nr)=n!r!(n−r)!\binom{n}{r} = \frac{n!}{r!(n-r)!} applies since the order of selection does not matter.
2
Determine the number of invalid combinations where both specific candidates are included.
(83)=8×7×63×2×1=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56
If both specific candidates are already placed on the panel, 3 more members must be chosen from the remaining 8 candidates.
3
Apply complementary counting to subtract invalid selections from total selections.
252−56=196252 - 56 = 196
Subtracting the restricted combinations from the total combinations gives the number of valid panel configurations.

Anahtar Kavram

Combinations with mutual exclusion (Complementary Counting)
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Konular

Combinations

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