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A committee of 55 members is to be selected from 66 doctors and 44 nurses. In how many ways can this committee be formed if it must include at least 33 doctors?

  1. 186Cevap
  2. B
    120
  3. C
    720
  4. D
    3600

Cevap

186 ways
The required committee must contain at least 3 doctors out of 5 total members. The three possible scenarios are: 3 doctors and 2 nurses (6C3×4C2=120^6C_3 \times ^4C_2 = 120), 4 doctors and 1 nurse (6C4×4C1=60^6C_4 \times ^4C_1 = 60), and 5 doctors and 0 nurses (6C5×4C0=6^6C_5 \times ^4C_0 = 6). Summing these gives 120+60+6=186120 + 60 + 6 = 186.

Adım Adım Çözüm

1
Identify the possible valid committee compositions given the condition 'at least 3 doctors'.
Three mutually exclusive cases exist for a 5-member committee: (3 doctors, 2 nurses), (4 doctors, 1 nurse), or (5 doctors, 0 nurses).
The total size of the committee is 5, so selecting more doctors reduces the required number of nurses.
2
Calculate the combinations for Case 1: 3 doctors and 2 nurses.
6C3×4C2=6×5×43×2×1×4×32×1=20×6=120^6C_3 \times ^4C_2 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{4 \times 3}{2 \times 1} = 20 \times 6 = 120
Order of selection within each group does not matter.
3
Calculate the combinations for Case 2: 4 doctors and 1 nurse.
6C4×4C1=6×52×1×4=15×4=60^6C_4 \times ^4C_1 = \frac{6 \times 5}{2 \times 1} \times 4 = 15 \times 4 = 60
Selecting 4 doctors out of 6 is equivalent to choosing which 2 doctors to exclude.
4
Calculate the combinations for Case 3: 5 doctors and 0 nurses.
6C5×4C0=6×1=6^6C_5 \times ^4C_0 = 6 \times 1 = 6
Choosing 5 doctors out of 6 yields 6 possibilities, and choosing 0 nurses yields 1.
5
Sum the number of ways from all valid cases.
120+60+6=186120 + 60 + 6 = 186
By the addition principle of counting, the total ways to form the committee is the sum of ways across mutually exclusive cases.

Anahtar Kavram

Combinations with constraints (addition and multiplication principles)
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