# Practice Paper 1 Question 13

On a grid of \(m\times n\) squares, how many ways exist to get from the top-left corner to the bottom-right corner if you can only move right or down on an edge?

## Related topics

## Warm-up Questions

16 tennis players participate in a

*doubles*tournament and are paired at random. How many ways are there to select a pair?How many different ways can a football team's manager choose 5 penalty takers from 11 players?

## Hints

- Hint 1How many steps in total must be taken?
- Hint 2From a total of \(m+n\) steps, how many are rightward steps and how many are downward steps?
- Hint 3If we have \(m\) rightward steps, how many combinations of them are there?
- Hint 4For each combination of rightward steps, how many combinations of downward steps are there?

## Solution

Regardless of the path taken, you must travel \(m\) steps right and \(n\) steps down, i.e. you must always travel \(m+n\) steps. Of these, we consider the number of different ways we may choose \(m\) to be rightward steps. This is \(\binom{m+n}{m}=\frac{(m+n)!}{n!m!}\). For each of these combinations, the remaining \(n\) steps down are uniquely determined i.e. there is only 1 combination since \(\binom{m+n-m}{n} = \binom{n}{n} = 1\), so \(\binom{m+n}{m}\) is our final answer.

Similarly, we may first find the combinations of \(n\) downward steps and fix rightward steps to obtain \(\binom{m+n}{n}\) as our answer.

Another way to look at this is to consider the permutations of \(m+n\) steps, where \(m\) rightward steps and \(n\) downward steps are both indistinguishable types of steps. We find the number of permutations for \(m+n\) things, and remove (by dividing by) the number of permutations specifically for just the rightwards and downward steps. This gives us the answer \(\frac{(m+n)!}{m! \cdot n!}\).

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