Chapter 4.5, Problem 13E

(a)

To determine

To find:

The number of different schedules possible if he needs to have an equal number of symphonies from Mozart, Haydn, and Schubert.

Answer to Problem 13E

The number of different schedules possible if he needs to have an equal number of symphonies from Mozart, Haydn, and Schubert is $1.2490697e+23$.

Explanation of Solution

Given:

In choosing music to play at a charity fund-raising event, Marlow has 41 Mozart, 104 Haydn, and 8 Schubert Symphonies from which to choose. He is setting up a schedule of the 12 songs to be played during the show.

Formula Used:

Formula to calculate the number of combinations,

$C{}_r=\frac{n!}{r!(n-r)!}$

Where, n is the number of distinct objects, and r is the number of objects selected from a group of n distinct objects.

Calculation:

12 songs can be played in the show from 41 Mozart, 104 Haydn and 8 Schubert symphonies. If an equal number of songs are to be chosen from Mozart, Haydn And Schubert, Marlow must choose 4 symphonies of Mozart, Haydn And Schubert.

4 Mozart symphonies out of 41 can be chosen:

$\begin{array}{rll}C{}_4& =& \frac{41!}{4!(41-4)!}\\ & =& \frac{41!}{4!37!}\\ & =& \frac{41\times \cancel{\stackrel{5}{40}}\times \cancel{\stackrel{13}{39}}\times 38\times \cancel{37!}}{(\cancel{4\times 3\times 2\times 1})\times \cancel{37!}}\\ & =& 101270\end{array}$

4 Haydn symphonies out of 104 can be chosen:

$\begin{array}{rll}C{}_4& =& \frac{104!}{4!(104-4)!}\\ & =& \frac{104!}{4!100!}\\ & =& \frac{\cancel{\stackrel{13}{104}}\times 103\times \cancel{\stackrel{34}{102}}\times 101\times \cancel{100!}}{(\cancel{4\times 3}\times \cancel{2}\times 1)\times \cancel{100!}}\\ & =& 4598126\end{array}$

4 Schubert symphonies out of 8 can be chosen:

$\begin{array}{rll}C{}_4& =& \frac{8!}{4!(8-4)!}\\ & =& \frac{8!}{4!4!}\\ & =& \frac{\cancel{8}\times 7\times \cancel{\stackrel{2}{6}}\times 5\times 4!}{(\cancel{4\times 3\times 2\times 1})\times \cancel{4!}}\\ & =& 560\end{array}$

Total number of arrangements:

$\begin{array}{rll}P{}_{12}& =& \frac{12!}{(12-12)!}\\ & =& \frac{12!}{0!}\\ & =& 12!\\ & =& 479001600\end{array}$

The number of different schedules possible if he needs to have an equal number of symphonies from Mozart, Haydn, and Schubert is:

$\begin{array}{rll}C{}_4\times C{}_4\times C{}_4\times P{}_{12}& =& 101270\times 4598126\times 560\times 479001600\\ & =& 1.2490697e+23\end{array}$ $$

Thus, the number of different schedules possible if he needs to have an equal number of symphonies from Mozart, Haydn, and Schubert is $1.2490697e+23$.

(b)

To determine

To find:

The probability that all 12 symphonies are by Mozart, if the songs are chosen randomly.

Answer to Problem 13E

The probability that all 12 symphonies are by Mozart, if the songs are chosen randomly is $3.579429631972963e-6$.

Explanation of Solution

Calculation:

There are 41 Mozart, 104 Haydn and 8 Schubert symphonies, and 12 songs are to be chosen randomly in $C$ ways:

$\begin{array}{rll}C{}_{12}& =& \frac{153!}{12!(153-12)!}\\ & =& \frac{153!}{12!141!}\\ & =& \frac{153\times 152\times 151\times 150\times 149\times 148\times 147\times 146\times 145\times 144\times 143\times 142\times 141!}{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)\times 141!}\\ & =& 220667975965944780\end{array}$,

if all 12 symphonies are by Mozart, it can be chosen in $C{}_{12}$.

$\begin{array}{rll}C{}_{12}& =& \frac{41!}{12!(41-12)!}\\ & =& \frac{41}{12!29!}\\ & =& \frac{41\times 40\times 39\times 38\times 37\times 36\times 35\times 34\times 33\times 32\times 31\times 30\times 29!}{(12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)\times 29!}\\ & =& 7898654920\end{array}$

The required probability is:

$\begin{array}{rll}\frac{C{}_{12}}{C{}_{12}}& =& \frac{7898654920}{220667975965944780}\\ & =& 3.579429631972963e-6\end{array}$

Thus, the probability that all 12 symphonies are by Mozart, if the songs are chosen randomly is $3.579429631972963e-6$.