A flea, which is at point $A$ on a line, jumps exactly $1 \mathrm{~m}$ each time, without ever leaving this line.

a) If the flea wants to reach point $B$ located on the line, $5 \mathrm{~m}$ to the right of $A$, with exactly 7 jumps, in how many ways can it do this?

b) If the flea wants to reach point $C$ located on the line, $5 \mathrm{~m}$ to the right of $A$, with exactly 9 jumps, in how many ways can it do this?

c) Is it possible for the flea to reach point $D$ located on the line, $2013 \mathrm{~m}$ from $A$, with exactly 2028 jumps? Justify.
Solution

a) To reach $B$, the flea must take exactly one step to the left and six to the right, in any order. Since this step to the left can be taken at any time, there are 7 possible moments to take it! Therefore, there are 7 distinct ways for the flea to reach $B$ in 7 steps.

b) To reach $C$, the flea must take 7 steps to the right and 2 to the left, in any order. How many ways can the flea do this? One way is to list them all, which are 36.

Another, more interesting way, is to think about it as follows: how many ways can we arrange 9 distinct objects? The answer is $9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$. And if there are 7 objects of one type and 2 of another type? Then, in the previous way, we would be counting each configuration many times. In fact, we would be counting $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ times due to one type of repeated object and $2 \times 1$ times due to the other. Therefore, we just need to do

$$
\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1) \times (2 \times 1)} = \frac{9 \times 8}{2 \times 1} = 36
$$

and thus, there are 36 ways. We can now think of an object of one type as being a jump to the right, and the object of the other type as being a jump to the left. Thus, each way of arranging the objects corresponds to a way of arranging the flea's jumps. We conclude that the flea has 36 distinct ways to reach $C$.

c) The answer is no! Because $2028 - 2013 = 15$, which is odd.