Tutorial Exercises

1

If $m>n$, then a one-to-one / injective function is not possible as the entire domain must be mapped so at least two elements in the domain will be mapped to the same one in the codomain.

Otherwise if $m\le n$, given an arbitrary $m$, say $3$, they could be mapped: $123$, $132$, $213$, $231$, $312$, $321$. In this case, order matters but repetition is not allowed.

Hence we use $n\cdot (n-1)\cdot ...\cdot (n-m+1)$.

2

todo

3

1. $3$ They can either take out two of the same one, or take out one then the other and then it doesn’t matter which they pick because they will have both.
2. $14$ They may take out all $12$ brown socks, then they need to take out an additional $2$ black socks.

title:
Correct answer.

4

1. $8$ different types but we are selecting $6$ where order matters but repetition is not allowed. $8-6+1=3$ $\frac{6!}{3!}=120$
2. Only one variety allowed at once, so $8$.
3. Up to $8$ different types, where doesn’t order matter and repetition is allowed. $\cancel{8^6=262,144}$ $\left(\begin{array}{c}6+8-1\\ 7\end{array}\right)=\left(\begin{array}{c}13\\ 7\end{array}\right)=1716$
4. As above but ignore where there is only one variety, so $1716-8=1708$.
5. The blueberry-filled donuts can be in $3$ positions out of $6$ where order doesn’t matter and repetition is not allowed. So $\left(\begin{array}{c}3+8-1\\ 7\end{array}\right)=120$
6. Number in (c) minus number in (e), which is $1716-120=1596$

5

For each $MMMMM$ or $WWWWW$ within the row: Order matters, repetition not allowed for any 5, so $k!=5!=120$. (for one set)

Now to apply it to both sets: $120\times 120=14,400$.

They may be placed $WMWMWMWMWM$ or $MWMWMWMWMW$, so $14,400\times 2=28,800$.

title:
Correct answer.

6

How many different words can be made by rearranging letters in $BANANA$?

• $B$ may be placed in any of the $6$ positions. ($6$)
• The three $A$s can be placed in any of the $5$ positions. ($\left(\begin{array}{c}5\\ 3\end{array}\right)$)
• The two $N$s will be placed in the last two remaining places.

$$\left(\begin{array}{cc}6\times \left(\begin{array}{c}5\\ 3\end{array}\right)& =6\times 10\\ & =60\text{ different words}\end{array}\right)$$

title:
Correct answer.

7

Prove Pascal’s identity using defining formula of $\left(\begin{array}{c}n\\ k\end{array}\right)$. (slides 129/127)