Question

if c(n,4),c(n,5) and c(m,6) are in AP, then what is the values of n________.

• A. 8
• B. 7
• C. 9
• D. 10

Answer 1

The Correct Option is B

To solve the problem, we start by considering the condition given: \(C(n,4)\), \(C(n,5)\), and \(C(m,6)\) are in arithmetic progression (AP). In an AP, the difference between consecutive terms is constant. Using the combination formula \(C(n,k) = \frac{n!}{k!(n-k)!}\), we have:

\(C(n,5) - C(n,4) = C(m,6) - C(n,5)\)

Now calculate each combination:
\(C(n,4) = \frac{n!}{4!(n-4)!} = \frac{n(n-1)(n-2)(n-3)}{24}\)
\(C(n,5) = \frac{n!}{5!(n-5)!} = \frac{n(n-1)(n-2)(n-3)(n-4)}{120}\)
\(C(m,6) = \frac{m!}{6!(m-6)!} = \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{720}\)

Substitute into the AP condition:
\(\frac{n(n-1)(n-2)(n-3)(n-4)}{120} - \frac{n(n-1)(n-2)(n-3)}{24} = \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{720} - \frac{n(n-1)(n-2)(n-3)(n-4)}{120}\)

Simplify the expressions:
Factor out \(\frac{n(n-1)(n-2)(n-3)}{24}\) from the left side:
\(= \frac{n(n-1)(n-2)(n-3)}{24}\left(\frac{n-4}{5} - 1\right) = \frac{n(n-1)(n-2)(n-3)}{24}\left(\frac{n-9}{5}\right)\)

Now the equation becomes:
\(\frac{n(n-1)(n-2)(n-3)(n-9)}{120} = \frac{m(m-1)(m-2)(m-3)(m-4)(m-5)}{720}\)

After evaluating the possibilities for \(n\), it turns out that when \(n = 7\), the condition holds, and both sides of the equation will be balanced.
Therefore, the value of \(n\) is 7.