Problem 1 (permutation)

• Five people, designed as A, B, C, D E, are arranged in linear order.

1. How many ways to arrange these five people?
   • $5!=120$ ways
2. How many ways to arrange these five people, if they are arranged in a circle?
   • as you can rotate the same combination into five different order:
   • $5!/5=24$ ways

Problem 2 (permutation)

• How many different ways can 3 red, 4 yellow and 2 blue bulbs be arranged in a string of Christmas tree lights with 9 sockets? $$\begin{array}{rl}\frac{9!}{4!3!2!}& =\frac{9\cdot 8\cdot 7\cdot 6\cdot 5}{3!2!}\\ & =\frac{9\cdot 8\cdot 7\cdot 5}{2!}\\ & =9\cdot 4\cdot 7\cdot 5\\ & =9\cdot 140\\ & =1260\end{array}$$
• 1260 ways?

Problem 3 (combinations)

• A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if

1. there are no restrictions?
   • $$\begin{gathered} \left(\genfrac{}{}{0ex}{}{10}{5}\right)=\frac{20!}{(10-5)!5!}=\frac{10\cdot 9\cdot 8\cdot 7\cdot 6}{5!} \\ =\frac{9\cdot 8\cdot 7\cdot 6}{3\cdot 4}=9\cdot 4\cdot 7=252 \end{gathered}$$
   • 252 ways
2. one particular person must be chosen on the committee?
   • $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=\frac{9!}{(9-4)!4!}=126$
3. there are to be 3 men and 2 women?
   • $\left(\genfrac{}{}{0ex}{}{6}{3}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{2}\right)=20\cdot 6=120$
   • 120 ways
4. there is to be a majority of women?
   • either: 3 woman or 4
   • 3 women: $\left(\genfrac{}{}{0ex}{}{6}{2}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{3}\right)=15\cdot 4=60$
   • 4 women: $\left(\genfrac{}{}{0ex}{}{6}{1}\right)\cdot \left(\genfrac{}{}{0ex}{}{4}{4}\right)=6\cdot 1=6$
   • 60+6=66 ways

Problem 4 (combinations)

• If 4 Maths books are selected from 6 different Maths books, and 3 English books are selected from 5 different English books, how many ways can the seven books be arranged on a shelf (linear order):

1. if there are no restrictions?
   • $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 7!=756000$
2. if 4 Maths books remain together?
   • $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 4!4!=18000$
   • one $4!$ for arranging 4 "groups" (3 individual English books, and 1 group of math books)
   • and another for order within the math books
3. if Maths and English books alternate?
   • $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 4!3!=21600$
4. if a Math book is at the beginning of the shelf?
   • $\left(\genfrac{}{}{0ex}{}{6}{4}\right)\cdot \left(\genfrac{}{}{0ex}{}{5}{3}\right)\cdot 4\cdot 6!=432000$