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Chapter 9: Problem 39

A patient with end-stage kidney disease has nine family members who are potential kidney donors. How many possible orders are there for a best match, a second-best match, and a third-best match?

Short Answer

Expert verified
There are 504 possible orders for a best match, a second-best match, and a third-best match.

Step by step solution

01

Understand Permutations

Permutations refer to the arrangement of items in specific order. When order of selection matters, we use permutations. Use the formula for permutations which is \(P(n,r) = \frac{n!}{(n-r)!}\), where 'n' is the total number of items and 'r' is the number of items to choose and arrange.
02

Identify 'n' and 'r'

In this exercise, 'n' is the total number of potential kidney donors (9 family members), and 'r' is the number of donors to select and arrange (best match, second-best match, and third-best match). So, 'n' is 9 and 'r' is 3.
03

Substitute 'n' and 'r' into the permutations formula

Substitute 'n' = 9 and 'r' = 3 into the permutations formula: \(P(n,r) = \frac{n!}{(n-r)!} = \frac{9!}{(9-3)!} = \frac{9!}{6!}\)
04

Solve the permutation

Solve for the permutation. Remember that '!' (factorial) means to multiply the number by every positive integer less than itself down to 1. Therefore: \(P(n,r) = \frac{9*8*7*6*5*4*3*2*1}{6*5*4*3*2*1} = 9*8*7 = 504\) So, there are 504 possible orders for a best match, a second-best match, and a third-best match.

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Most popular questions from this chapter

Finding a Sum In Exercises \(45-54\) , find the sum using the formulas for the sums of powers of integers. $$\sum_{n=1}^{20}\left(n^{3}-n\right)$$

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