Question

Suppose you have 5 riders and 5 horses, and you want to pair them off so that every rider is assigned one horse (and no horse is assigned two riders). How many ways can you do this?

Answer 1

There are 120 ways in which 5  riders and 5 horses can be arranged.

We have,

5 riders and 5 horses,

Now,

We know that,

Now,

Using the arrangement formula of Permutation,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$,

So,

For n = 5,

And,

r = 5

As we have,

n = r,

So,

Now,

Using the above-mentioned formula of arrangement,

i.e.

The total number of ways $^nN_r = \frac{n!}{(n-r)!}$,

Now,

Substituting values,

We get,

$^5N_5 = \frac{5!}{(5-5)!}$

We get,

The total number of ways of arrangement = 5! = 5 × 4 × 3 × 2 × 1 = 120,

So,

There are 120 ways to arrange horses for riders.

Hence we can say that there are 120 ways in which 5  riders and 5 horses can be arranged.

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