At a blood drive, 5 donors with type O + blood, 5 donors with type A + blood, and 2 donors with type B + blood are in line. In how many distinguishable ways can the donors be in line? The donors can be in line in different ways. ...
At a blood drive, 5 donors with type O + blood, 5 donors with type A + blood, and 2 donors with type B + blood are in line. In how many distinguishable ways can the donors be in line? The donors can be in line in different ways. ...
Chapter9: Sequences, Probability And Counting Theory
Section9.5: Counting Principles
Problem 42SE: In horse racing, a “trifecta” occurs when a bettor wins by selecting the first three finishers in...
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![**Permutations of Donors with Different Blood Types**
At a blood drive, there are 5 donors with type O+ blood, 5 donors with type A+ blood, and 2 donors with type B+ blood queued up. We need to determine in how many distinguishable ways these donors can be arranged in line.
To solve this problem, we use the formula for permutations of a multiset:
\[ \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]
Where:
- \( n \) is the total number of items,
- \( n_1 \), \( n_2 \), and \( n_3 \) are the counts of each distinct type of item.
Given data:
- Total number of donors (\( n \)) = 5 (O+) + 5 (A+) + 2 (B+) = 12.
- \( n_1 \) (O+ donors) = 5.
- \( n_2 \) (A+ donors) = 5.
- \( n_3 \) (B+ donors) = 2.
Thus, the number of distinguishable ways the donors can be arranged in line is calculated as:
\[ \frac{12!}{5! \cdot 5! \cdot 2!} \]
Place the computed number in the following box:
The donors can be in line in \[ \boxed{} \] different ways.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F20a0be3e-3983-4420-aed0-1c62c53b9b01%2F7edd05f6-6036-4906-b2e5-58b5a1b74655%2Fvs2yfq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Permutations of Donors with Different Blood Types**
At a blood drive, there are 5 donors with type O+ blood, 5 donors with type A+ blood, and 2 donors with type B+ blood queued up. We need to determine in how many distinguishable ways these donors can be arranged in line.
To solve this problem, we use the formula for permutations of a multiset:
\[ \frac{n!}{n_1! \cdot n_2! \cdot n_3!} \]
Where:
- \( n \) is the total number of items,
- \( n_1 \), \( n_2 \), and \( n_3 \) are the counts of each distinct type of item.
Given data:
- Total number of donors (\( n \)) = 5 (O+) + 5 (A+) + 2 (B+) = 12.
- \( n_1 \) (O+ donors) = 5.
- \( n_2 \) (A+ donors) = 5.
- \( n_3 \) (B+ donors) = 2.
Thus, the number of distinguishable ways the donors can be arranged in line is calculated as:
\[ \frac{12!}{5! \cdot 5! \cdot 2!} \]
Place the computed number in the following box:
The donors can be in line in \[ \boxed{} \] different ways.
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