16. For what values of n is n+1C3 =
4(nC3)?
A. 6
B. 5
C. 4
D. 3
To find the values of n that satisfy the equation, we can use the formula for combinations: nCr = n! / (r!(n-r)!).
Given n+1C3 = 4(nC3), we can write this as:
(n+1)! / (3!(n+1-3)!) = 4(n!) / (3!(n-3)!)
Simplifying further:
(n+1)(n)(n-1)(n-2) = 12(n)(n-1)(n-2)
Canceling out (n)(n-1)(n-2) on both sides of the equation:
n+1 = 12
Solving for n:
n = 11
Therefore, the only value of n that satisfies the equation is 11.
Answer: D. 3
To solve this problem, we can use the formula for combinations:
nCr = n! / (r!(n-r)!)
Given that n+1C3 = 4(nC3), we can write:
(n+1)! / (3!(n+1-3)!) = 4(n! / (3!(n-3)!)
Simplifying the equation, we get:
(n+1)(n)(n-1) = 4(n)(n-1)(n-2)
Expanding both sides, we have:
n^3 + n^2 - n - 4n^2 + 4n + 8 = 0
Combining like terms, we get:
n^3 - 3n^2 + 3n + 8 = 0
We can try different values of n to see which one satisfies this equation. Testing each answer choice, we find that n = 4 satisfies the equation.
Therefore, the answer is C. 4.