Factorial notation problem. Please help?

Hi, I am stuck on this maths problem. Solve for n: (n+2)!/(n)! =110. I get as far as (n+2)(n+1) = 110. When I multiply it out I get n^2,3n,2 = 110. There is only one example similar to this in my book and it used a quadratic equation to solve it. I feel like that is the next step. I am just unsure of what to do with the 2 I have left over from multiplying out the binomial. I'm really sorry if this doesn't make sense. It's been several years since I had to study mathematics and I am really rusty. Any help is greatly appreciated.

110 = 2*5*11

So, if n=9, (n+1)(n+2) = 10*11 = 110

Or, algebraically,

n^2+3n+2 = 110
n^2+3n-108 = 0
(n+12)(n-9) = 0
n = 9,-12

Yes I think it does make sense because

n^2+3n+2 = 110
n^2+3n-108 = 0
(n+12)(n-9) = 0
n = 9,-12 makes all of it come together ❤

No problem! I can help you with this.

You are on the right track. From (n+2)(n+1) = 110, you can simplify it further by expanding the left side of the equation:

n^2 + 3n + 2 = 110

To solve this quadratic equation, we can rearrange it to:

n^2 + 3n + 2 - 110 = 0

Combine like terms:

n^2 + 3n - 108 = 0

Now, let's factorize the quadratic equation. We are looking for two numbers that multiply to give -108 and add up to 3.

The factors of -108 that add up to 3 are 12 and -9. Therefore, we can rewrite the equation as:

(n + 12)(n - 9) = 0

Set each factor equal to zero and solve for n:

n + 12 = 0 or n - 9 = 0

n = -12 or n = 9

So, the solutions for n are -12 and 9.

Keep in mind that since we are dealing with a factorial, n must be a non-negative integer. Therefore, the only valid solution is n = 9.

I hope this helps! Let me know if you have any further questions.

No problem at all! I'm here to help you with your math problem.

To solve the equation (n+2)! / (n)! = 110, we can begin by expanding the factorial notation. The factorial of a number n is denoted by n! and it represents the product of all positive integers from 1 to n.

So, we rewrite the equation as (n+2)(n+1)(n!) / (n)! = 110.

Now, we can cancel out the common factors of (n)! in the numerator and denominator, leaving us with (n+2)(n+1) = 110.

To solve this quadratic equation, we start by simplifying the equation further. Let's multiply out the binomial in the left-hand side: (n+2)(n+1) = n^2 + 3n + 2 = 110.

Next, we rearrange the equation to set it equal to zero: n^2 + 3n + 2 - 110 = 0.

Now, we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Since the coefficients in this equation are manageable, we can try factoring it.

To factor, we need to find two numbers whose product is 2 (the constant term) and whose sum is 3 (the coefficient of n). The numbers that satisfy these conditions are 2 and 1.

Thus, n^2 + 3n + 2 = (n + 2)(n + 1) = 0.

Setting each factor equal to zero, we get two possible solutions: n + 2 = 0, which gives n = -2, and n + 1 = 0, which gives n = -1.

Therefore, the solutions to the equation (n+2)! / (n)! = 110 are n = -2 and n = -1.

Remember, when dealing with factorial notation, the values of n must be non-negative integers. However, in this case, we obtained negative solutions because of the algebraic manipulation.

Since the negative solutions don't satisfy the non-negative integer requirement, there are no valid solutions for n in this particular equation.

I hope this explanation helps you understand how to approach and solve the factorial notation problem. If you have any further questions, feel free to ask!