Finding n and r from nPr and nCr

We are given the values for a permutation and a combination involving the same numbers, n and r:

• \({}^n P_r = 336\)
• \({}^n C_r = 56\)

We need to find the values of n and r.

Relationship between Permutations and Combinations

There is a direct relationship between the number of permutations (\({}^n P_r\)) and the number of combinations (\({}^n C_r\)) for the same values of n and r. The relationship is given by the formula:

$$ {}^n P_r = {}^n C_r \times r! $$

where \(r!\) is the factorial of r.

Calculating the Value of r

We can use the relationship above and the given values to find r!.

Substitute the given values into the formula:

$$ 336 = 56 \times r! $$

Now, solve for \(r!\):

$$ r! = \frac{336}{56} $$

Performing the division:

$$ r! = 6 $$

We need to find the value of r whose factorial is 6. Let's calculate the factorials of small integers:

• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)

Since \(3! = 6\), the value of r must be 3.

Calculating the Value of n

Now that we have the value of r (which is 3), we can use either the formula for \({}^n P_r\) or \({}^n C_r\) to find n. Let's use the formula for \({}^n P_r\):

$$ {}^n P_r = \frac{n!}{(n-r)!} $$

Substitute the known values: \({}^n P_3 = 336\). So,

$$ 336 = \frac{n!}{(n-3)!} $$

The expression \(\frac{n!}{(n-3)!}\) can be expanded as:

$$ \frac{n!}{(n-3)!} = \frac{n \times (n-1) \times (n-2) \times (n-3)!}{(n-3)!} = n \times (n-1) \times (n-2) $$

So, we have the equation:

$$ n \times (n-1) \times (n-2) = 336 $$

We are looking for three consecutive integers whose product is 336. We can test values for n, keeping in mind that \(n \ge r\), so \(n \ge 3\).

• If \(n=6\), \(6 \times 5 \times 4 = 120\) (Too small)
• If \(n=7\), \(7 \times 6 \times 5 = 210\) (Too small)
• If \(n=8\), \(8 \times 7 \times 6 = 336\) (Correct)

Thus, the value of n is 8.

Verification

Let's verify our values \(n=8\) and \(r=3\) using the original formulas:

For \({}^n P_r\):

$$ {}^8 P_3 = \frac{8!}{(8-3)!} = \frac{8!}{5!} = \frac{8 \times 7 \times 6 \times 5!}{5!} = 8 \times 7 \times 6 = 336 $$

This matches the given \({}^n P_r\) value.

For \({}^n C_r\):

$$ {}^8 C_3 = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} = \frac{8 \times 7 \times 6 \times 5!}{(3 \times 2 \times 1) \times 5!} = \frac{8 \times 7 \times 6}{6} = 8 \times 7 = 56 $$

This matches the given \({}^n C_r\) value.

Both values are consistent with the given information.

Conclusion

Based on the calculations, the values of n and r are 8 and 3, respectively.

Therefore, \(n = 8\) and \(r = 3\).