Section Solutions for Chapter 3
Solutions to Exercise 3.1.9
3.1.9.1 The band may choose any one of the \(\displaystyle 6\) people to play lead guitar. They may then choose any one of the remaining \(\displaystyle 5\) people to play bass. Therefore, the band can be completed in \(\displaystyle 6\cdot 5=30\) ways. You might also observe that the number of ways to complete the band is the number of \(\displaystyle 2\)-permutations (for the two open spots) of \(\displaystyle 6\) people, which is \(\displaystyle 6 \cdot \ldots \cdot (6-2+1)=6\cdot 5=30\text{.}\)
3.1.9.2 Divide this into two cases, depending on which of the two parts Garth got. If they got the first part, then there were \(\displaystyle 8\) other people competing for \(\displaystyle 4\) other roles, so the number of ways of completing the cast in this case is the number of \(\displaystyle 4\)-permutations of the \(\displaystyle 8\) people. Repeating this argument for the second case (if Garth got the other part) and adding the two numbers, we see that in total there are \(\displaystyle 2 \cdot 8\cdot 7 \cdot 6 \cdot 5\) (since \(\displaystyle 8-4+1=5\)) ways of completing the cast. This works out to \(\displaystyle 3,360\text{.}\)
Solutions to Exercise 3.2.10
3.2.10.2 At the end of this trick, the only sets of cards that they could not possibly end up with are sets that contain nothing but spades. There are \(\displaystyle \binom{13}{3}\) sets that include only spades (choose any 3 of the 13 spades), and \(\displaystyle \binom{52}{3}\) possible sets of 3 cards from the deck as a whole, so the number of sets of three cards that are not all spades is \(\displaystyle \binom{52}{3}-\binom{13}{3}=22,100-286=21,814\text{.}\)
3.2.10.3 The leading digit cannot be a zero, so if there are to be exactly two zeroes, we have \(\displaystyle 4\) possible positions in which they can be placed. Thus, there are \(\displaystyle \binom{4}{2}\) ways of choosing where to place the two zeroes. In each of the remaining three positions, we can place any of the digits \(\displaystyle 1\) through \(\displaystyle 9\text{,}\) so there are \(\displaystyle 9^3\) choices for the remaining digits. Thus, there are \(\displaystyle \binom{4}{2}9^3=4,374\) \(\displaystyle 5\)-digit numbers that contain exactly two zeroes.
Solutions to Exercise 3.3.7
3.3.7.2 Using the Binomial Theorem, we see that
\begin{equation*}
(a+b)^5(c+d)^6=\left(\sum_{r=0}^5 \binom{5}{r}a^rb^{5-r}\right)\left(\sum_{s=0}^6 \binom{6}{s}c^sd^{6-s}\right)\text{.}
\end{equation*}
To find the coefficient of \(\displaystyle a^2b^3c^2d^4\text{,}\) we must take \(\displaystyle r=2\) and \(\displaystyle s=2\text{.}\) This gives us the term \(\displaystyle \binom{5}{2}a^2b^3\binom{6}{2}c^2 d^4=\binom{5}{2}\binom{6}{2}a^2b^3c^2d^4\text{.}\) Thus, the coefficient of \(\displaystyle a^2b^3c^2d^4\) is \(\displaystyle \binom{5}{2}\binom{6}{2}=10\cdot 15=150\text{.}\)3.3.7.4 Using the Binomial Theorem, we see that
\begin{equation*}
(a+b)^5+(a+b^2)^4=\sum_{r=0}^5 \binom{5}{r}a^rb^{5-r}+\sum_{s=0}^4 \binom{4}{s}a^s(b^2)^{4-s}\text{.}
\end{equation*}
The coefficient of \(\displaystyle a^3b^2\) in the first summand arises when \(\displaystyle r=3\text{;}\) in the second summand, it arises when \(\displaystyle s=3\text{.}\) This gives us the term \(\displaystyle \binom{5}{3}a^3b^2+\binom{4}{3}a^3(b^2)^1\text{.}\) Thus, the coefficient of \(\displaystyle a^3b^2\) is \(\displaystyle \binom{5}{3}+\binom{4}{3}=10+4=14\text{.}\)