# Solve the following equation for n C (n+1, 3) = C(n,2) I know the answer is 5 - since I did trial and error, but how would I show the work?

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### 2 Answers

You'll solve this equation involving combinations using the factorial formula for combinations of n elements taken k at a time:

C(n,k) = n!/k!(n - k)!

Let's evaluate C(n+1 , 3) = (n+1)!/3!(n+1-3)! = (n+1)!/3!(n - 2)!

Let's evaluate C(n , 2) = n!/2!(n-2)!

Now, we'll equate the equivalent expressions:

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

We'll simplify both sides by (n-2)!

(n+1)!/3! = n!/2!

We can write (n+1)! = n!*(n+1)

We can write 3! = 2!*3

n!*(n+1)/2!*3 = n!/2!

We'll simplify both sides by n!/2!:

(n+1)/3 = 1

We'll cross multiply:

n + 1 = 3

n = 3 - 1

n = 2

**Since the value of "n" has to be a natural number, therefore a positive integer, we'll accept n=2 as solution of the given equation.**

the answer is NOT 5!

**C**(a,b) may also be written as a**C**b, implying a!/b!(a-b)!

Hence, (n+1)**C**3 = (n+1)!/3!(n+1-3)!

But (n+1)! = (n+1)n!

And (n+1-3)! = (n-2)!

Therefore,

(n+1)**C**3**=** (n+1)n!/3!(n-2)!

Also, n**C**2 = n!/2!(n-2)!

Thus, (n+1)n!/3!(n-2)! = n!/2!(n-2)!

At this step, multiply both sides of the equation with (n-2)!/n!

This leaves us with the equation:

(n+1)/3! = 1/2!

Multiply both sides by 3!, giving us:

n+1 = 3!/2!

Hence, n+1 = 3

And n = 3-1;

**n = 2**