If (n+1)C(n-1) = 28, find the value of n

C stands for Combinations

but n^2 + n - 56 = (n+8)(n-7)

Thanks Steve, good catch.

(n+1)C(n-1) = 28

(n+1)! / ((n+1-n+1)! (n-1)! ) = 28
(n+1)! / (2! (n-1)! ) = 28
(n+1)(n)(n-1)!/(2! (n-1)! ) = 28
n(n+1)/2 = 28
n^2 + n = 56
n^2 + n- 56 = 0
(n-8)(n+7) = 0
n = 8 or n = -7 , but n ≥ 0

n = 8

another way would be to realize that n+1 and n-1 have a difference of 2.
So in your Pascal's Triangle go down in the third column to find 28
1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 ....

look at the number in front of 28

Thanks!!

To solve the equation (n + 1)C(n - 1) = 28, we can first understand what the expression means.

The C, in this case, represents the binomial coefficient, also known as a combination. It calculates the number of ways to choose a certain number of items from a larger set. The notation (n + 1)C(n - 1) means "n + 1 choose n - 1."

To simplify this expression, we can use the formula for combinations:

(n + 1)C(n - 1) = (n + 1)! / [(n - 1)! * 1!]

The exclamation mark (!) represents the factorial of a number, which means multiplying all positive integers up to that number. For example, 5! = 5 * 4 * 3 * 2 * 1.

Expanding the expression further:

(n + 1)! / [(n - 1)! * 1!] = (n + 1) * n! / [(n - 1)! * 1] = (n + 1) * n! / (n - 1)!

Now, simplify the expression by canceling out the common terms:

(n + 1) * n! / (n - 1)! = (n + 1) * n

So our equation becomes (n + 1) * n = 28.

To find the value of n, we can rearrange the equation:

n^2 + n - 28 = 0

This equation is a quadratic equation which we can solve for n by factoring, completing the square, or using the quadratic formula.

Factoring the equation, we have:

(n - 4)(n + 7) = 0

This gives us two possible solutions:

n - 4 = 0 or n + 7 = 0

Therefore, n = 4 or n = -7.

However, since n represents the number of items chosen, we eliminate the negative solution n = -7 because it doesn't make sense in this context.

Therefore, the value of n is 4.