For what of n is n+1C3 = 4(nC3)?
Question 1 Report
For what of n is n+1C3 = 4(nC3)?
Answer Details
To solve the given equation, we can use the formula for combinations, which is:
nCr = n! / r!(n-r)!
Substituting n+1 for n and 3 for r, we get:
(n+1)C3 = (n+1)! / 3!(n+1-3)!
(n+1)C3 = (n+1)! / 6n!
Simplifying the right side of the equation using the given information, we get:
4(nC3) = 4n! / 6n!
(nC3) = n! / 6n!
Now we have:
(n+1)C3 = 4(nC3)
Substituting the formulas we derived earlier, we get:
(n+1)! / 3!(n+1-3)! = 4n! / 3!(n-3)!
Simplifying and canceling the terms, we get:
(n+1)(n)(n-1) = 4(n)(n-1)(n-2)
Expanding and simplifying the equation, we get:
n^3 - 3n^2 - 10n + 12 = 0
Factoring the equation, we get:
(n-4)(n^2 - n - 3) = 0
The roots of the equation are n = 4, (1+sqrt(13))/2, and (1-sqrt(13))/2. Since n represents a counting number, the only possible solution is n = 4.
Therefore, the answer is 4.
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