At a blood drive, 4 donors with type 0+ blood, 3 donors with type A+ blood, and 3 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, 4 donors with type 0+ blood, 3 donors with type A+ blood, and 3 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.
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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![**Problem Statement:**
At a blood drive, 4 donors with type O+ blood, 3 donors with type A+ blood, and 3 donors with type B+ blood are in line. In how many distinguishable ways can the donors be in line?
**Solution Explanation:**
To find the number of distinguishable ways the donors can be arranged, we use the formula for permutations of a multiset:
\[
\frac{n!}{n_1! \times n_2! \times n_3!}
\]
Where:
- \( n \) is the total number of donors.
- \( n_1, n_2, n_3 \) are the counts of each type.
In this case:
- Total donors, \( n = 4 + 3 + 3 = 10 \)
- Donors with type O+ blood, \( n_1 = 4 \)
- Donors with type A+ blood, \( n_2 = 3 \)
- Donors with type B+ blood, \( n_3 = 3 \)
Plugging in the values:
\[
\frac{10!}{4! \times 3! \times 3!} = \frac{3,628,800}{24 \times 6 \times 6} = 12,600
\]
The donors can be in line in **12,600** different ways.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74bd7c97-b14c-43b6-a405-6d68de03e70c%2F8ca8039c-6c79-477d-a108-0490195cd1cc%2Ffjj6nzx_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
At a blood drive, 4 donors with type O+ blood, 3 donors with type A+ blood, and 3 donors with type B+ blood are in line. In how many distinguishable ways can the donors be in line?
**Solution Explanation:**
To find the number of distinguishable ways the donors can be arranged, we use the formula for permutations of a multiset:
\[
\frac{n!}{n_1! \times n_2! \times n_3!}
\]
Where:
- \( n \) is the total number of donors.
- \( n_1, n_2, n_3 \) are the counts of each type.
In this case:
- Total donors, \( n = 4 + 3 + 3 = 10 \)
- Donors with type O+ blood, \( n_1 = 4 \)
- Donors with type A+ blood, \( n_2 = 3 \)
- Donors with type B+ blood, \( n_3 = 3 \)
Plugging in the values:
\[
\frac{10!}{4! \times 3! \times 3!} = \frac{3,628,800}{24 \times 6 \times 6} = 12,600
\]
The donors can be in line in **12,600** different ways.
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