To find the values of n and r, we are given two equations:

1) nPr = 336
2) nCr = 56

Let's solve these equations step by step.

Finding nPr = 336:
nPr represents the number of permutations of r objects taken from a set of n objects. The formula for nPr is given by:

nPr = n! / (n - r)!

Where "!" denotes the factorial of a number, which means multiplying all positive integers from 1 to that number.

To find nPr = 336, we need to find the values of n and r that satisfy this equation. We can start by listing down the factorials of numbers until we find a combination that yields 336.

1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320, 9! = 362880, ...

Looking at the factorials, we see that 8! is the closest factorial to 336. We can use this information to narrow down our search.

8! = 40320

Now, we need to find the value of r. We can rewrite the equation nPr = 336 as:

n! / (n - r)! = 336

Substituting the value of 8! for n!, we get:

40320 / (n - r)! = 336

Simplifying further, we have:

(n - r)! = 40320 / 336
(n - r)! = 120

Looking at the factorials again, we see that 5! is the closest factorial to 120.

5! = 120

Therefore, we can conclude that n - r = 5.

Now, let's move on to the second equation.

Finding nCr = 56:
nCr represents the number of combinations of r objects taken from a set of n objects. The formula for nCr is given by:

nCr = n! / (r! * (n - r)!)

Using the same process as before, we can rewrite the equation nCr = 56 as:

n! / (r! * (n - r)!) = 56

Substituting the value of 5! for (n - r)!, we get:

n! / (r! * 5!) = 56

Now, we have two equations:

1) n - r = 5
2) n! / (r! * 5!) = 56

Using these two equations, we can solve for the values of n and r by substitution or trial and error.

By trying different values of r, we can find that when r = 3, n = 8 satisfies both equations.

So, the values of n and r are n = 8 and r = 3.

Therefore, n = 8 and r = 3 satisfy the given conditions nPr = 336 and nCr = 56.