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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?

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Warm-up Questions

1. 16 tennis players participate in a doubles tournament and are paired at random. How many ways are there to select a pair?

2. How many different ways can a football team's manager choose 5 penalty takers from 11 players?

Hints

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