Test 26 (Lessons 51–52): Counting Principles Solutions

1. Construct rows $0–7$ of Pascal’s Triangle.

2. Expand ${(2x–y)}^4$ using Pascal’s Triangle.

$$\begin{gathered} \mathrm{Row} 4:\hspace{0.17em}1, 4, 6, 4, 1 \\ 1{\left(2x\right)}^4{\left(–y\right)}^0+4{\left(2x\right)}^3{\left(–y\right)}^1+6{\left(2x\right)}^2{\left(–y\right)}^2+4{\left(2x\right)}^1{\left(–y\right)}^3+1{\left(2x\right)}^0{\left(–y\right)}^4 \\ {2}^4{x}^4+4·2^3{x}^3{\left(–y\right)}^1+6·2^2{x}^2{\left(–y\right)}^2+4·2x{\left(–y\right)}^3+{\left(–y\right)}^4 \\ 16x^4–32x^{3}y+24x^2{y}^2–8x{y}^3+y^4 \end{gathered}$$

3. $nCr(6, 4)$

15

4. $nCr(7, 5)$

21

5. You stop at the grocery store to purchase five items. How many different ways can you take the items out of your basket to purchase?

$5!=5·4·3·2·1=120$
There are 120 arrangements of the five items.

6. There are twelve students in the debate club at West High School. Their advisor needs to assign them to committees. How many ways can the club members be arranged into committees of three?

$\begin{array}{rcl}nCr\left(12, 3\right)& =& \frac{12!}{3!\left(12–3\right)!}=\frac{12·11·10·\cancel{9!}}{3! \cancel{9!}}\\ & =& \frac{{\displaystyle \stackrel{2}{\cancel{12}}}·11·10}{\cancel{3·2·1}}=2·11·20=220\end{array}$

The students can be arranged into 220 different committees of three.

7. In a tournament, the first- and second-place teams receive awards. How many arrangements are possible if sixteen teams sign up to compete?

$\begin{array}{rcl}nPr\left(16, 2\right)& =& \frac{16!}{\left(16–2\right)!}=\frac{16·15·\cancel{14!}}{\cancel{14!}}\\ & =& 16·15\\ & =& 240\end{array}$

There are 240 ways (permutations) that sixteen teams can place first and second.

8. Find the fifth term of the binomial: ${\left(\frac{1}{2}b+c\right)}^9$

$\begin{array}{ccc}nCr\left(9, 4\right)=\frac{9!}{4!\left(9–4\right)!}=126& & n=9, r=5–1=4\\ 126{\left(\frac{1}{2}b\right)}^5{\left(c\right)}^4& & \\ 126·\frac{1}{32}{b}^5{c}^4& & 9–4=5\\ \frac{63}{16}{b}^5{c}^4& & \end{array}$

9. Find the middle term of the binomial: ${\left(x–y\right)}^{14}$

$$\begin{gathered} nCr\left(14, 7\right)=\frac{14!}{7!\left(14–7\right)!}=3432 \\ 3432{\left(x\right)}^7{\left(–y\right)}^7 \\ –3432x^7{y}^7 \end{gathered}$$

10. You read twenty books this year and decided to write a review of your top five books.
    How many ways can you rank your top five books of the year?

$\begin{array}{rcl}nPr\left(20, 5\right)& =& \frac{20!}{\left(20–5\right)!}\\ & =& 20·19·18·17·16\\ & =& 1,860,480\end{array}$

There are 1,860,480 permutations by which to rank your top five books of the year.