16. For what values of n is n-1C3 =

4(nC3)?
A. 6
B. 5
C. 4
D. 3

To solve this problem, we need to express the combinations in terms of factorials.

The formula for nCr is given by nCr = n! / (r!(n-r)!).

Using this formula, we can express n-1C3 as (n-1)! / (3!((n-1)-3)!), which simplifies to (n-1)! / 3!(n-4)! = (n-1)(n-2)(n-3) / 3!.

Similarly, nC3 can be expressed as n! / (3!(n-3)!), which simplifies to n(n-1)(n-2) / 3!.

We are given that n-1C3 = 4(nC3), so we can set up the equation:

(n-1)(n-2)(n-3) / 3! = 4(n)(n-1)(n-2) / 3!

Simplifying, we have:

(n-3) = 4n

n - 3 = 4n

-3 = 3n

n = -1

However, since we are looking for values of n that make the equation valid, we must check if n = -1 satisfies the equation or not.

Plugging in n = -1, we have:

(-1 - 3) = 4(-1)

-4 = -4

Since the equation is true, n = -1 is a valid solution.

Therefore, the answer is n = -1.

The correct option is not given in the answer choices.

To find the values of n that satisfy the equation n-1C3 = 4(nC3), we can use the formula for combinations.

First, let's expand the terms using the combination formula:

(n-1)C3 = ((n-1)!)/((3!)(n-1-3)!) = ((n-1)!)/(3!(n-4)!)

4(nC3) = 4(n!)/((3!)(n-3)!)

Now we can equate the two expressions:

((n-1)!)/(3!(n-4)!) = 4(n!)/((3!)(n-3)!)

Next, we can simplify the equation by canceling out common factors:

(n-1)!/(n-4)! = 4(n!)/(n-3)!

Now, let's focus on canceling out the factorials. We can notice that the factorials in both fractions can be canceled out up to the (n-4) term:

(n-1)(n-2)(n-3)(n-4)!/(n-4)! = 4n(n-1)(n-2)(n-3)/(n-3)(n-4)!

(n-1)(n-2)(n-3) = 4n(n-1)(n-2)

Now, let's simplify the equation further:

(n-1)(n-2)(n-3) - 4n(n-1)(n-2) = 0

Now, we can start canceling out the common factors again:

(n-1)(n-2)[(n-3)-4n] = 0

(n-1)(n-2)(-3n-3) = 0

Now, we have three factors that could be equal to zero:

1. (n-1) = 0
2. (n-2) = 0
3. (-3n-3) = 0

Solving these equations, we get:

1. n-1 = 0
n = 1

2. n-2 = 0
n = 2

3. -3n-3 = 0
-3n = 3
n = -1

Therefore, the values of n that satisfy the equation n-1C3 = 4(nC3) are n = 1, n = 2, and n = -1.

Answer: There is no option provided in the given question that includes all three of the possible values for n.