Question

integrated math
final review - ch 4-1 topics
you may use this review as notes on the final.

1. use a calculator to compute the following:

₁₃p₂ =
₁₀c₉ =
4! =
₁₀p₉ =
₁₀c₄ =
9! =

1. tsyrn has 11 books, how many ways can they choose 5 to bring on his camping trip? (they can read them in any order later)
2. a padlock is numbered 1-40. how many 3-digits codes are possible if numbers cannot be repeated?
3. a bike lock has 6 rollers numbered 0-9. how many codes are possible? (numbers can be repeated)
4. someone has 8 different plants to place in her window. how many ways can she arrange them?

Explanation:

Step 1: Calculate \(_{13}P_7\)

The permutation formula is \(_{n}P_{r}=\frac{n!}{(n - r)!}\). For \(n = 13\) and \(r=7\), we can also use a calculator. \(_{13}P_7=13\times12\times11\times10\times9\times8\times7 = 8648640\) (using calculator: \(13\) permute \(7\) is \(8648640\))

Step 2: Calculate \(_{16}C_9\)

The combination formula is \(_{n}C_{r}=\frac{n!}{r!(n - r)!}\). For \(n = 16\) and \(r = 9\), \(_{16}C_9=\frac{16!}{9!(16 - 9)!}=\frac{16!}{9!7!}=\frac{16\times15\times14\times13\times12\times11\times10}{7\times6\times5\times4\times3\times2\times1}=11440\) (using calculator: \(16\) choose \(9\) is \(11440\))

Step 3: Calculate \(4!\)

\(n!=n\times(n - 1)\times\cdots\times1\), so \(4!=4\times3\times2\times1 = 24\)

Step 4: Calculate \(_{15}P_8\)

Using permutation formula or calculator, \(_{15}P_8=\frac{15!}{(15 - 8)!}=15\times14\times13\times12\times11\times10\times9\times8=259459200\)

Step 5: Calculate \(_{10}C_4\)

Using combination formula or calculator, \(_{10}C_4=\frac{10!}{4!(10 - 4)!}=\frac{10\times9\times8\times7}{4\times3\times2\times1}=210\)

Step 6: Calculate \(9!\)

\(9!=9\times8\times7\times6\times5\times4\times3\times2\times1 = 362880\)

Step 7: Solve part 2 (Taryn's books)

This is a combination problem since order does not matter. \(n = 11\), \(r=5\). \(_{11}C_5=\frac{11!}{5!(11 - 5)!}=\frac{11!}{5!6!}=\frac{11\times10\times9\times8\times7}{5\times4\times3\times2\times1}=462\)

Step 8: Solve part 3 (Padlock code)

This is a permutation problem (order matters, no repetition). \(n = 40\), \(r = 3\). \(_{40}P_3=40\times39\times38=59280\)

Step 9: Solve part 4 (Bike lock code)

Each roller has 10 options (0 - 9) and there are 6 rollers. Since repetition is allowed, the number of codes is \(10\times10\times10\times10\times10\times10=10^6 = 1000000\)

Step 10: Solve part 5 (Arranging plants)

Assuming the number of plants is \(n\) (the text seems to have a typo, but if we assume \(n\) plants, the number of arrangements is \(n!\). If we assume \(n = 8\) as per the partial text "8 different plants"), then \(8!=40320\) (if \(n\) is different, adjust accordingly, but with the given text, we assume \(n = 8\))

Answer:

1. \(_{13}P_7=\boldsymbol{8648640}\), \(_{16}C_9=\boldsymbol{11440}\), \(4!=\boldsymbol{24}\), \(_{15}P_8=\boldsymbol{259459200}\), \(_{10}C_4=\boldsymbol{210}\), \(9!=\boldsymbol{362880}\)
2. \(\boldsymbol{462}\) ways
3. \(\boldsymbol{59280}\) possible codes
4. \(\boldsymbol{1000000}\) possible codes
5. If 8 plants, \(\boldsymbol{40320}\) ways (assuming 8 plants from the text "8 different plants")