**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.

Answered: Image /qna-images/answer/8ca8039c-6c79-477d-a108-0490195cd1cc.jpg

Answered: At a blood drive, 4 donors with type 0+…

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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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.