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Chapter 2: Q33E (page 73)

Again, consider a Little League team that has \({\rm{15}}\) players on its roster.

a. How many ways are there to select \({\rm{9}}\) players for the starting lineup?

b. How many ways are there to select \({\rm{9}}\)players for the starting lineup and a batting order for the \({\rm{9}}\) starters?

c. Suppose \({\rm{5}}\) of the \({\rm{15}}\) players are left-handed. How many ways are there to select \({\rm{3}}\) left-handed outfielders and have all \({\rm{6}}\) other positions occupied by right-handed players?

Short Answer

Expert verified

a) \({\rm{1,816,214,400}}\)

b) \({\rm{659,067,881,472,000;}}\)

c) \({\rm{9,072,000}}\)

Step by step solution

01

team roaster consists of

A team's roster consists of 15 players.

02

determining ways to select 9 players for the starting lineup 

We utilize permutations since the order is crucial (we have distinct positions) and we have \({\rm{15}}\) players and \({\rm{9}}\) positions for players.

An ordered subset is referred to as a permutation. The number of permutations of size \({\rm{k}}\) for \({\rm{n}}\)persons in a group is represented as \({{\rm{P}}_{{\rm{k,n}}}}\).The proposal is this:

\({{\rm{P}}_{{\rm{k,n}}}}{\rm{ = }}\frac{{{\rm{n!}}}}{{{\rm{(n - k)!}}}}\)

We have some possibilities too,

\({{\rm{P}}_{{\rm{9,15}}}}{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{(15 - 9)!}}}}{\rm{ = }}\frac{{{\rm{15!}}}}{{{\rm{6!}}}}{\rm{ = 1,816,214,400}}\)

03

 Determining the ways are there to select 9 players for the starting lineup and a batting order for the 9 starters

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

The ordered pair in this case is a starting lineup in first place and a batting order in second place. We can pick a starting lineup in \({\rm{1,816,214,400}}\)ways (a), and we can choose the batting order in \({\rm{1,816,214,400}}\)ways

\({{\rm{P}}_{{\rm{9,9}}}}{\rm{ = }}\frac{{{\rm{9!}}}}{{{\rm{(9 - 9)!}}}}{\rm{ = }}\frac{{{\rm{9!}}}}{{{\rm{1!}}}}{\rm{ = 362,880}}\)

We need permutations of size \({\rm{n = 9}}\)for \({\rm{k = 9}}\)people. The ultimate solution, based on everything we've discussed, is

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 1,816,214,400*362,880 = 659,067,881,472,000}}{\rm{.}}\)

04

Determining the ways to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players

The sequence is still important, so there are a few options.

\({{\rm{P}}_{{\rm{3,5}}}}{\rm{ = }}\frac{{{\rm{5!}}}}{{{\rm{(5 - 3)!}}}}{\rm{ = 60}}\)

three left-handed athletes to pick from and

\({{\rm{P}}_{{\rm{6,10}}}}{\rm{ = }}\frac{{{\rm{10!}}}}{{{\rm{(10 - 6)!}}}}{\rm{ = 151,200}}\)

strategies to choose six out of ten left-handed athletes for the remaining six spots From:

If we assume that we may pick the first element of an ordered pair in \({{\rm{n}}_{\rm{1}}}\)ways and the second element in \({{\rm{n}}_{\rm{2}}}\)ways for each selected element, then the number of pairings is \({{\rm{n}}_{\rm{1}}}{{\rm{n}}_{\rm{2}}}\)

\({{\rm{n}}_{\rm{1}}}{\rm{*}}{{\rm{n}}_{\rm{2}}}{\rm{ = 60*151,200 = 9,072,000}}{\rm{.}}\)

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Most popular questions from this chapter

The composer Beethoven wrote \({\rm{9}}\) symphonies, \({\rm{9}}\) piano concertos (music for piano and orchestra), and \({\rm{32}}\) piano sonatas (music for solo piano).

a. How many ways are there to play first a Beethoven symphony and then a Beethoven piano concerto?

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a. \(P\left( {{A_2}|{A_1}} \right)\)

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d. \(P\left( {{A_1} \cap {A_2} \cap {A_3}|{A_1} \cup {A_2} \cup {A_3}} \right)\)

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