Siddharth's Paths

Since we want the grasshopper to reach the line $x = 36$ , we could ignore the ordinates of its motion. Hence, the problem is reduced to finding the number of ways for the grasshopper to jump $6$ or $9$ steps to the right, which is similar to finding the number of orderings of a sum of $6s$ and $9s$ to form $36$ . We break the problem into cases.

(To make it clear, the top number of the binomial coefficient represents the number of addends; the bottom number gives the number of $6s$ in the sum. This could be done, WLOG, for the number of $9s$ , as the numbers appear repeatedly in the sum, and the binomial coefficients will give similar results.)

CASE 1: Grasshopper jumps from $(m, n)$ to $((m + 6), (n + 9))$ at all times.

This is similar to ordering $\frac {36} {6} = 6$ $6s$ , which could be done in $\binom {6} {6} = \frac {6!} {6! 0!} = 1$ way.

CASE 2: Grasshopper jumps from $(m, n)$ to $((m + 6), (n + 9))$ at some instances, OR to $((m + 9), (n + 6))$ at others.

This is similar to ordering however many $6s$ and $9s$ can be fitted to make $36$ . But the LCM of $9$ and $6$ is $18$ ; hence, $3$ $6s$ can be replaced by $2$ $9s$ , and indeed $36 = 2 \times 9 + 3 \times 6$ . For this case, $3$ $6s$ and $2$ $9s$ can be ordered in $\binom {5} {3} = \frac {5!} {3! 2!} = 10$ ways.

CASE 3: Grasshopper jumps from $(m, n)$ to $((m + 9), (n + 6))$ at all times.

We now use $\frac {36} {9} = 4$ $9s$ , which could be done in $\binom {4} {0} = \frac {4!} {0! 4!} = 1$ way.

Therefore, the grasshopper can jump from the origin to the line $x = 36$ in $\boxed {1 + 10 + 1 = 12}$ ways.

Let the grasshopper takes a steps of type-1 and b steps of type-2.

Therefore, 6a + 9b = 36.

a = 6, b = 0 (1 way)

a = 3, b = 2 (in any order, hence 10 ways)

a = 0, b = 4 (1 way). Hence 12 different paths.

The question can be posed as: "How many different ways can you go up a flight of stairs of length 36 steps taking only 6 or 9 steps at a time?". Dividing all the values by 3, the question becomes "How many different ways can you go up a flight of stairs of length 12 steps taking only 2 or 3 steps at a time?".

This is a recursion problem, in which F (n+3) = F (n+1) + F (n), where F (n) is the number of ways to reach step "n".

Using this formula, there are 0 ways to get to the 1st step, 1 way to get to the 2nd step, 1 way to get to the 3rd step, (0+1=1) way to get to the 4th step, (1+1=2) ways to get to the 5th step, and so on so forth.

The number of ways to reach the 1st - 12th steps in order is: 0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12 Thus, our answer is 12.

We can reduce the problem to using jumps of (m+3, n+2) and (m+2, n+3) to get to x = 12. Using binomial coefficients and counting possible paths, we get 1+ 10+ 1 = 12 possible ways.

Let's just look at the m part of the options, as we only care about how far the grasshopper is x-wise.

We then define the end possibilities: all 6s (six 6s) or all 9s. (four 9s). We also know that two 9s will cover the same distance as three 6s. Thus there are three possible endtotals of moves: all 6s, all 9s, or three 6s two 9s.

However, we are counting paths. There is only one possible path for each end scenario, so we only have to use combinations to find the number of arrangements for the middle one. 5c3 = 10.

10 + 2 = 12.

The grasshopper jumps from the point (m, n) to (m+6, n+9) or (m+9, n+6). In other words, in each move grasshopper advances 9 or 6 units in x-direction and similarly in y–direction. Since, the grasshopper is at (0, 0) at the beginning and want to go to line x=36. The problem reduces to the number of ways grasshopper can cover these 36 units. Number of ways of doing this = breaking 36 into steps of 6 and 9 and arranging them (i) 4 steps of 9 each = 1 (ways) (ii) 2 steps of 9 and 3 steps of 6 =Fact(5)/[Fact(3)*Fact(2)] = 10 (ways) (iii) 6 steps of 6 each = 1 (ways) Total number ways of doing this = 1 + 10 + 1 = 12

For this, we only need to consider the jumps from m. There are three ways to arrive exactly an line x=36: 1. Use six (m+6) jumps. There is only one way of achieving this. 2. Use four (m+9) jumps. There is also only one way of doing this. 3. Use three (m+6) and two (m+9) jumps. We have 5 jumps for this. Considering that we treat each jump uniquely, we have 5! possible jump combinations. However, of these 5 jumps, we have three of the same kind and two of the other kind. Thus, for this case, we have (5!)/(3!2!) = 10 possible jump combinations. Adding them all up, we have 12 ways to get to line x=36.

If you don't know how to simply your calculation, a possibility that works is to write down the "possibilities tree", where you see all the possible ways.

Obs.: There are only 5 steps, so it's a valid method if you think in the time of resolution.

Since we are only interested in getting to a given $x$ , we can disregard the $y$ component of jumps. Notice that in this case it is possible to get to $x = 36$ with

1. 6 jumps with $x$ component 6
2. 3 jumps with $x$ component 6 and 2 jumps with $x$ component 9
3. 4 jumps with $x$ component 9

In cases 1 and 3, there is obviously only one path. In case 2, the number of unique paths is given by $\frac{5!}{2!3!} = 10$ . Therefore, the answer is 12.

Initially $m=0$ and the question essentially asks: if every second you either add a $6$ or a $9$ then how many ways can you get $36$ ?

Two ways are to have six $6's$ or to have four $9's$ .

Since $36$ and $6$ are both even, we can see that if we were to have some $6's$ and some $9's$ then there must be an even number of $9's$ .

The only even number less than $4$ but more than $0$ is $2$ which would mean we need three $6's$

Since the order of the $9's$ and $6's$ matter there are $\frac{5!}{3! * 2!} = 10$ ways to have a chain of $6's$ and $9's$ that add up to $36$

Therefore there are a total of $1+1+10 = 12$ ways for the grasshopper to land on the line $x=36$

let x be the number of step (6;9) and y be the number of step (9,6) To reach the line x= 36, x,y must satisfy 6x+9y=36. The solution are (x,y)=(0,4),(3,2),(6,0) <> (x,y)=(0,4), (6,0). there is one way of arrangement. <> (x,y)=(3,2). The number of possible ways is \frac{(2+3)!}{2!*3!} = 10 therefore, the total number of paths is 10+1+1=12

since since grasshoper has to go to x=36 possible combinations with the above situation= 6+6+6+6+6+6 .............1 combination 9+9+9+9......................1combination 6+6+6+9+9...................10 combinations hence there are 12 ways

We'll just look at the $x$ coordinates. We know that the sum of $x$ coordinates should be $36$ . So as per the conditions given, we can say $6q+9r=36$ $$ This diophantine equation in non-negative integers has solutions $r=0,2,4$ $$ Now in any path, there are either zero nines, or two nines or four nines as $x$ coordinates (We may ignore the $m$ and $n$ as the starting point is $(0,0)$ ) $$ Now if there are two nines in the path, number of ways = ${5}\choose{2}$ = $10$ $$ If there are zero nines or four nines, there is only $1$ possible way in each case. $$ So our final answer is $\boxed {1+1+10=12}$

We only care about the first component $m$ . We can add to it either $9$ or $6$ in order to get 36. The $9$ 's come in pairs because they are odd have to add up to an even number. This leaves 3 possible sequences: $6-6-6-6-6-6, 9-9-6-6-6, 9-9-9-9$ . There is only one permutation to the first and last. The middle one has $\frac{5!}{2!3!} = 10$ permutations as given by the binomial coefficient. Therefore the answer is $1 + 1 + 10 = 12$ .

Note that $36 = (4)(9) = (6)(6) = 18 + 18 = (3)(6) + (2)(9)$ . We have two cases to consider. First, the grasshopper takes only one kind of jump, either 4 steps of the form $(m+9,n+6)$ or 6 steps of the form $(m+6,n+9)$ . That gives us 2 paths. If the grasshopper takes more than one kind of jump, then a moment of reflection shows us that it must take 2 steps of the form $(m+9,n+6)$ and 3 of the other. The number of paths like this is $\binom{5}{2} = 10$ . This gives us a total of 12.

let the grashopper jumps to $m+6$ $a$ times and jumps to $m+9)$ $b$ times so the grashopper land on $6a+9b$ if we want the grashopper land on $x=36$ so $6a+9b=36$ it easy to check $(a,b)=(0,4),(6,0),(3,2)$ for $(a,b)=(0,4),(6,0)$ it have one path each one. for $(a,b)=(3,2)$ there are $\frac{5!}{3!2!}=10$ so there are 12 different path for grashopper land on the line $x=36$

If the grasshopper only used the first method of jumping, (m+6,n+9), to reach x=36, there is 1 possible path. If the grasshopper only used the second method of jumping, (m+9,n+6), to reach x=36, there is also only 1 possible path. If the grasshopper used both methods to reach x=36, there are 4+3+2+1=10 or 4(4+1)/2=4(5)/2=20/2=10 possible paths. Adding all the solutions up you get 1+1+10=12 total paths.

There is exactly one way to reach the line $x=36$ using only the $(m+9, n+6)$ jump (let's call this the "6-jump", after $n+6$ ), and only one way to do the same using only the $(m+6, n+9)$ jump (the "9-jump"). Also, there are $\frac {5\times4} {2} = 10$ ways to reach the line using three 6-jumps and two 9-jumps; it can be verified that these are the only sequences that land on the required line. Hence there are a total of $12$ paths.

By the way, I hope this explanation is clear enough, I made it in a hurry.