This problem requires us to find the value of C(n + 1, r + 1) given the values of P(n, r) and C(n, r). We need to use the relationship between permutations and combinations to find the values of n and r first.

Finding the Value of r using Permutations and Combinations

The relationship between the number of permutations P(n, r) and the number of combinations C(n, r) for selecting r items from a set of n distinct items is given by the formula:

\( P(n, r) = C(n, r) \times r! \)

Where r! is the factorial of r.

We are given:

• \( P(n, r) = 2520 \)
• \( C(n, r) = 21 \)

Substitute the given values into the formula:

\( 2520 = 21 \times r! \)

To find r!, we divide P(n, r) by C(n, r):

\( r! = \frac{2520}{21} \)

\( r! = 120 \)

Now we need to find the integer r whose factorial is 120. Let's calculate the factorials of small integers:

• \( 1! = 1 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• \( 4! = 4 \times 3 \times 2 \times 1 = 24 \)
• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Since \( 5! = 120 \), we find that \( r = 5 \).

Determining the Value of n using Combinations

Now that we know \( r = 5 \) and \( C(n, r) = 21 \), we can use the combination formula to find the value of n.

The formula for combinations is:

\( C(n, r) = \frac{n!}{r! (n-r)!} \)

Substitute the values \( r = 5 \) and \( C(n, 5) = 21 \):

\( C(n, 5) = \frac{n!}{5! (n-5)!} = 21 \)

We know \( 5! = 120 \). So, the equation becomes:

\( \frac{n!}{120 (n-5)!} = 21 \)

Multiply both sides by 120:

\( \frac{n!}{(n-5)!} = 21 \times 120 \)

\( \frac{n!}{(n-5)!} = 2520 \)

Recall that \( \frac{n!}{(n-r)!} = P(n, r) \). So, we have \( P(n, 5) = 2520 \). This confirms our initial given information.

\( P(n, 5) = n \times (n-1) \times (n-2) \times (n-3) \times (n-4) = 2520 \)

We need to find an integer n such that the product of n and the next four consecutive decreasing integers is 2520. Let's test values for n starting from a number slightly greater than r (which is 5):

• If \( n = 6 \), \( P(6, 5) = 6 \times 5 \times 4 \times 3 \times 2 = 720 \) (Too small)
• If \( n = 7 \), \( P(7, 5) = 7 \times 6 \times 5 \times 4 \times 3 = 2520 \) (Correct)

So, we find that \( n = 7 \).

Calculating C(n + 1, r + 1)

We have found that \( n = 7 \) and \( r = 5 \). We are asked to find the value of \( C(n + 1, r + 1) \).

Substitute the values of n and r:

\( C(n + 1, r + 1) = C(7 + 1, 5 + 1) = C(8, 6) \)

Now, we calculate \( C(8, 6) \) using the combination formula \( C(n, r) = \frac{n!}{r! (n-r)!} \):

\( C(8, 6) = \frac{8!}{6! (8-6)!} \)

\( C(8, 6) = \frac{8!}{6! 2!} \)

Expand the factorials. Note that \( 8! = 8 \times 7 \times 6! \):

\( C(8, 6) = \frac{8 \times 7 \times 6!}{6! \times (2 \times 1)} \)

Cancel out the \( 6! \) term from the numerator and denominator:

\( C(8, 6) = \frac{8 \times 7}{2 \times 1} \)

\( C(8, 6) = \frac{56}{2} \)

\( C(8, 6) = 28 \)

Thus, the value of C(n + 1, r + 1) is 28.

Revision Table: Permutations and Combinations

Additional Information: Properties of Combinations

Understanding properties of combinations can be helpful in solving problems.

• \( C(n, r) = C(n, n-r) \)
• \( C(n, 0) = 1 \)
• \( C(n, n) = 1 \)
• \( C(n, 1) = n \)
• Pascal's Identity: \( C(n, r) + C(n, r+1) = C(n+1, r+1) \)

In this problem, we calculated \( C(8, 6) \). Using the property \( C(n, r) = C(n, n-r) \), we could also calculate \( C(8, 6) \) as \( C(8, 8-6) = C(8, 2) \).

\( C(8, 2) = \frac{8!}{2! (8-2)!} = \frac{8!}{2! 6!} = \frac{8 \times 7 \times 6!}{2 \times 1 \times 6!} = \frac{56}{2} = 28 \)

This gives the same result and demonstrates a useful property.