Permutations & Combinations Question - None

Question

Given ${}^nC_4$, ${}^nC_5$ and ${}^nC_6$ are in A.P., then $2\cdot{}^nC_5 = {}^nC_4 + {}^nC_6$. Find the value of $n$ (take the larger value).

Accepted Answer

Step 1: Write the A.P. condition using the given equation:

2·ⁿC₅ = ⁿC₄ + ⁿC₆

Step 2: Express combinations in factorial form:

2·[n!/(5!(n-5)!)] = n!/(4!(n-4)!) + n!/(6!(n-6)!)

Step 3: Factor out n!/(6!(n-6)!) and simplify:

2·[6(n-5)/(6·5)] = [(6·5)/(1)] + 1

2·[(n-5)/5] = 30 + 1 = 31

Step 4: Alternatively, divide the entire equation by ⁿC₅:

2 = (ⁿC₄/ⁿC₅) + (ⁿC₆/ⁿC₅)

Using ⁿCᵣ₊₁/ⁿCᵣ = (n-r)/(r+1):

2 = (5/(n-4)) + ((n-5)/6)

Step 5: Multiply through by 6(n-4):

12(n-4) = 30 + (n-5)(n-4)

12n - 48 = 30 + n² - 9n + 20

0 = n² - 21n + 98

Step 6: Using the quadratic formula or factoring:

n² - 21n + 98 = 0

(n - 14)(n - 7) = 0

n = 14 or n = 7

∴ The larger value is n = 14

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