Cha-cha-cha

The man has to walk $10$ block-lengths up and $10$ block-lengths to the right; he has to take a path (right, up, right, up, etc.), where each section is at least $2$ blocks long, So, we list out the partitions of $10$ (ways to represent $10$ as the sum of positive integers) with each part at least $2,$ divided up by the number of parts:

1. $1$ partition with $1$ part: $\begin{aligned} 11 &\text{ x1} \end{aligned}$

2. $7$ partitions with $2$ parts: $\begin{aligned} 2+8 &\text{ x2}, \\ 3+7 &\text{ x2}, \\ 4+6 &\text{ x2}, \\ 5+5 &\text{ x1} \end{aligned}$

3. $15$ partitions with $3$ parts: $\begin{aligned} 2+2+6 &\text{ x3}, \\ 2+3+5 &\text{ x6}, \\ 2+4+4 &\text{ x3}, \\ 3+3+4 &\text{ x3} \end{aligned}$

4. $10$ partitions with $4$ parts: $\begin{aligned} 2+2+2+4 &\text{ x4}, \\ 2+2+3+3 &\text{ x6} \end{aligned}$

5. $1$ partition with $5$ parts: $\begin{aligned}2+2+2+2+2 &\text{ x1}\end{aligned}$

(The multipliers after each partition are to adjust for ordering.)

Since the man starts by going right, the path must have either same number of right-parts as up-parts (for example, right-up-right-up), or $1$ more right-part than up-parts (for example, right-up-right-up-right). Also, once the number of right-parts and up-parts is determined, any $2$ partitions with those numbers of parts will work. Using these facts, we list out all possible cases:

1. $1$ part right, $1$ part up - there are $1 \times 1 = 1$ paths.
2. $2$ parts right, $1$ part up - there are $7 \times 1 = 7$ paths.
3. $2$ parts right, $2$ parts up - there are $7 \times 7 = 49$ paths.
4. $3$ parts right, $2$ parts up - there are $15 \times 7 = 105$ paths.
5. $3$ parts right, $3$ parts up - there are $15 \times 15 = 225$ paths.
6. $4$ parts right, $3$ parts up - there are $10 \times 15 = 150$ paths.
7. $4$ parts right, $4$ parts up - there are $10 \times 10 = 100$ paths.
8. $5$ parts right, $4$ parts up - there are $1 \times 10 = 10$ paths.
9. $5$ parts right, $5$ parts up - there are $1 \times 1 = 1$ paths.

Thus, in total, there are $1+7+49+105+225+150+100+10+1=\boxed{648}$ possible paths the man could take. $\blacksquare$

We start with the following result: For positive integers $n$ and $m$ , the number of solutions to $a_1 + a_2 + \dots + a_n = m,$ where each $a_i$ is a positive integer that is at least 2, is $\binom{m - n - 1}{n - 1}$ .

Proof : Let $b_i = a_i - 1$ , so each $b_i$ is a positive integer. Then $a_i = b_i + 1$ . Substituting and simplifying, we get $b_1 + b_2 + \dots + b_n = m - n.$ By stars-and-bars, the number of solutions to this equation is $\binom{m - n - 1}{n - 1}.$ It is easy to see that every solution in the $b_i$ corresponds to a solution in the $a_i$ . $\blacksquare$

We can represent each path as a string of Rs and Us, where each R corresponds to a step to the right, and each U corresponds to a step up. This string will contain 10 Rs and 10 Us. Furthermore, by the specifications in the problem, each "block" of Rs and Us will contain at least two letters. (For example, the string RRRUURRRRRUUUU consists of the four blocks RRR, UU, RRRRR, and UUUU.)

Let $r$ be the number of blocks of Rs, and let $u$ be the number of blocks of Us. Since the first step is to the right, and the blocks must alternate between Rs and Us, either $r = u$ or $r = u + 1$ .

Consider the case $r = u$ . Each block of Rs must contain at least two Rs, and there are 10 Rs total, so $1 \le r \le 5$ . By the result we proved at the start, there are $\binom{9 - r}{r - 1}$ ways to distribute 10 Rs among $r$ blocks, and since $u = r$ , we have the same number for the Us.

Next, consider the case $r = u + 1$ . In this case, $2 \le r \le 5$ . Again, there are $\binom{9 - r}{r - 1}$ ways to distribute 10 Rs among $r$ blocks, and since $u = r - 1$ , there are $\binom{9 - u}{u - 1} = \binom{10 - r}{r - 2}$ ways to distribute the Us.

Therefore, the total number of possible paths is $\sum_{r = 1}^5 \binom{9 - r}{r - 1} \binom{9 - r}{r - 1} + \sum_{r = 2}^5 \binom{9 - r}{r - 1} \binom{10 - r}{r - 2} = 648.$

There is also a solution using recursion. Let $f(r,u)$ denote the number of strings containing $r$ Rs and $u$ Us, so that each block contains at least two letters. Then $f$ satisfies the recursion $f(r,u) = f(r - 1, u) + f(r, u - 1) - f(r - 1, u - 1) + f(r - 2, u - 2)$ for all $r \ge 2$ and $u \ge 2$ , except when $r = u = 2$ . (I will leave the proof of this as an exercise.)

Here are the first few values of $f(r,u)$ :

$\begin{array}{c|ccccc} r \backslash u & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline 0 & 1 & 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 0 & 2 & 2 & 3 & 4 \\ 3 & 1 & 0 & 2 & 2 & 3 & 4 \\ 4 & 1 & 0 & 3 & 3 & 6 & 9 \\ 5 & 1 & 0 & 4 & 4 & 9 & 14 \end{array}$

Continuing the table, we find $f(10,10) = 1296$ , so the answer is $1296/2 = 648$ . (We divide by 2, because in the problem, the first step is to the right.)

Simple exhaustion method:

The man can take the paths

$2-2-2-2-2$ ,

$2-2-2-4, 2-2-3-3$ ,

$2-2-6, 2-3-5, 2-4-4, 3-3-4$ ,

$2-8, 3-7, 4-6, 5-5$ ,

$10$ .

With order into account, this makes $1, 10, 15, 7, 1$ paths of length $5, 4, 3, 2, 1$ , respectively. He starts with the right; if he takes a $n$ step path in the right direction then he must take either a $n$ or $n-1$ step path in the up direction. Thus, the sum is $1(10 + 1) + 10(10 + 15) + 15(15 + 7) + 7(7+1) + 1(1) = 648$ .

So first we note that any path must either have the same number of horizontal and vertical segments, or one more horizontal segment. Each solution will be a pairing of a two partitions of 10 into parts of at least 2. Now let's count the partitions of 10 into parts of at least 2, grouping by number of parts:

1 part: 1

2 parts:7

3 parts:15

4 parts:10

5 parts:1

Now we square each of these and add them to get $1+7*7+15*15+10*10+1=376$ Now we add each times the one after for the case where the number of segments are not equal: $1*7+7*15+15*10+10*1=272$ . 272+376=648

Each way is made of alternating R (right) and U (up) 's starting with R. For each such way number of R's would decide which partition of 10 we need to use. Same applies for U's.

Example: If we have RURUR as path pattern, then we will use 3 part for R's and 2 part for U's.

Ways to distribute 10 into smaller integers such that each part is greater than or equal to 2 is:

Into 1 part : - $10$

Into 2 part : - $2+8 \space , \space 3+7 \space , \space 4+6 \space , \space 5+5$

Into 3 part : - $2+2+6 \space , \space 2+3+5 \space , \space 2+4+4 \space , \space 3+3+4$

Into 4 part : - $2+2+2+4 \space , \space 2+2+3+3$

Into 5 part : - $2+2+2+2+2$

Number of ways partitions can be used can be calculated as the number of different permutations of that particular partitions

Example: 2 Part can have 2 permutation each of $( 2+8 \space , \space 3+7 \space , \space 4+6 )$ and 1 permutation of $5+5$ . Thus total of 7 permutations.

We get,

1 Part $= 1$

2 Part $= 7$

3 Part $= 15$

4 Part $= 10$

5 Part $= 1$

To get the number of ways any particular pattern can be traveled we simply need to multiply R's part ways to U's part ways, since they are mutually independent.

Example: RURUR can be traveled in $(3 Part) \times (2 Part) = 15 \times 7 = 105 \space ways$ .

Now counting the number of ways the man can travel along particular path pattern:

RU = $1 \times 1 = 1$

RUR = $7 \times 1 = 7$

RURU = $7 \times 7 = 49$

RURUR = $15 \times 7 = 105$

RURURU = $15 \times 15 = 225$

RURURUR = $10 \times 15 = 150$

RURURURU = $10 \times 10 = 100$

RURURURUR = $1 \times 10 = 10$

RURURURURU = $1 \times 1 = 1$

Adding up we get $\boxed{648}$

Look at the possible partitions of $10$ using numbers only greater than or equal to $2$ :

$10,8+2,7+3,6+4,5+5,\\6+2+2,5+3+2,4+4+2,4+3+3,\\4+2+2+2,3+3+2+2,2+2+2+2+2$

But since we also have to consider permutations of these partitions, there is $1$ partition with length $1$ , $7$ with length $2$ , $15$ with length $3$ , $10$ with length $4$ , and $1$ with length $5$ . A possible path can either have two with the same length or one with length $1$ less than the other one, so the answer is

$1+7(7+1)+15(15+7)+10(10+15)+1(1+10)=\boxed{648}$