Permutation Calculation

The problem asks to find the value of $r$ given the permutation equation ${^{10}\text{P}_r = 5040}$.

Permutation Formula

Recall the formula for permutations:

$^{n}\text{P}_r = \frac{n!}{(n-r)!}$

Where $n$ is the total number of items, and $r$ is the number of items to choose and arrange.

Applying the Formula

In this case, $n = 10$. Substituting the given values into the formula:

$^{10}\text{P}_r = \frac{10!}{(10-r)!} = 5040$

Solving for r

We need to find the value of $r$ such that the product of $r$ terms starting from 10 downwards equals 5040.

Let's expand the permutation:

• If $r=1$, $^{10}\text{P}_1 = 10$.
• If $r=2$, $^{10}\text{P}_2 = 10 \times 9 = 90$.
• If $r=3$, $^{10}\text{P}_3 = 10 \times 9 \times 8 = 720$.
• If $r=4$, $^{10}\text{P}_4 = 10 \times 9 \times 8 \times 7 = 5040$.

We found that when $r=4$, the value matches the given ${^{10}\text{P}_r}$.

Alternatively, using the factorial expansion:

$\frac{10!}{(10-r)!} = 5040$

We know $10! = 3,628,800$. Checking the value $5040$:

$10 \times 9 \times 8 \times 7 = 5040$

Comparing $10 \times 9 \times 8 \times 7$ with $\frac{10!}{(10-r)!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{ (10-r)! }$, we see that for the equation to hold true, we must have $(10-r)! = 6!$.

This implies $10-r = 6$.

Solving for $r$:

$r = 10 - 6$

$r = 4$

Conclusion

The value of $r$ is 4.