4. Solve for n if P(n, 3) = 6P(n – 1, 3). FOR FULL MARKS , you need to show the algebra not just state the answer by trail and error.
3 years ago
Answered By Emily D
This is a messy one, feel free to ask more questions if it doesn't make sense!
The reason it's messy is there are no real solutions for your equation. I've seen this question before as P(n-1, 2) which would give you an answer of n = 6. Is it possible you made a typo? If that's the case, feel free to skip the next paragraph and follow along with your numbers instead of the numbers I'm using!
The way your equation is written out now, we're left with: n = 1, 2, or 3.6
As you know, we cannot traditionally take factorials of negative numbers (1 - 3)!, zero (1 - 1)!, or decimals (3.6)!. Since this is a MATH 251 problem, you may have been introduced to factorials using decimals and negative numbers through approximation with the Gamma function, so I'll go through and show you how to solve it that way.
P(n,r) = n! / (n-r)!
So let's rewrite this equation in terms of n's and r's to get rid of the difficult-to-read notation:
n! / (n-3)! = 6(n-1)! / ((n-1)-3)!
Now let's take a look at the long way of writing out factorials:
n! = n * (n-1) * (n-2) * ... * (2) * (1)
What this tells us is that we can pull out as many beginning terms as we want from each of the factorials until they match each other. First I'm going to clean up our right-hand denominator: ((n-1)-3) = (n-4)
n! / (n-3)! = 6(n-1)! / (n-4)!
==> Now I'm going to pull a term out of our left denominator's factorial
n! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> we'll pull a term from our left numerator's factorial
n(n-1)! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> taking a side-step into weird, we'll keep pulling terms out of the two numerators; this will net us our first two solutions: n = 1 or n = 2
==> We cannot go further, because n = 3 would mean we are dividing by zero and our equation would be undefined! This is why n = 3, 4, etc is not a solution
==> now that we have those two, let's rewind to where the equation was less messy:
n(n-1)! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> (n-1)! is in both numerators and we're assuming it's a non-zero value. This is why the solutions of n = 1 is extra weird but, with those assumptions we can divide both sides by (n-1)!
n / [(n-3)*(n-4)!] = 6 / (n-4)!
==> eliminate (n-4)! from the denominator of both sides
n / (n-3) = 6
==> multiply both sides by n-3
n = 6(n-3)
==> expand the right-hand side
n = 6n - 18
==> move like-terms to the same side
5n = 18
==> solve for n by dividing 18/5 to get our final solution
3 years ago
Answered By Emily D
This is a messy one, feel free to ask more questions if it doesn't make sense!
The reason it's messy is there are no real solutions for your equation. I've seen this question before as P(n-1, 2) which would give you an answer of n = 6. Is it possible you made a typo? If that's the case, feel free to skip the next paragraph and follow along with your numbers instead of the numbers I'm using!
The way your equation is written out now, we're left with: n = 1, 2, or 3.6
As you know, we cannot traditionally take factorials of negative numbers (1 - 3)!, zero (1 - 1)!, or decimals (3.6)!. Since this is a MATH 251 problem, you may have been introduced to factorials using decimals and negative numbers through approximation with the Gamma function, so I'll go through and show you how to solve it that way.
P(n,r) = n! / (n-r)!
So let's rewrite this equation in terms of n's and r's to get rid of the difficult-to-read notation:
n! / (n-3)! = 6(n-1)! / ((n-1)-3)!
Now let's take a look at the long way of writing out factorials:
n! = n * (n-1) * (n-2) * ... * (2) * (1)
What this tells us is that we can pull out as many beginning terms as we want from each of the factorials until they match each other. First I'm going to clean up our right-hand denominator: ((n-1)-3) = (n-4)
n! / (n-3)! = 6(n-1)! / (n-4)!
==> Now I'm going to pull a term out of our left denominator's factorial
n! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> we'll pull a term from our left numerator's factorial
n(n-1)! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> taking a side-step into weird, we'll keep pulling terms out of the two numerators; this will net us our first two solutions: n = 1 or n = 2
n(n-1)(n-2)(n-3)(n-4)! / [(n-3)*(n-4)!] = 6(n-1)(n-2)(n-3)(n-4)!/(n-4)!
==> We cannot go further, because n = 3 would mean we are dividing by zero and our equation would be undefined! This is why n = 3, 4, etc is not a solution
==> now that we have those two, let's rewind to where the equation was less messy:
n(n-1)! / [(n-3)*(n-4)!] = 6(n-1)! / (n-4)!
==> (n-1)! is in both numerators and we're assuming it's a non-zero value. This is why the solutions of n = 1 is extra weird but, with those assumptions we can divide both sides by (n-1)!
n / [(n-3)*(n-4)!] = 6 / (n-4)!
==> eliminate (n-4)! from the denominator of both sides
n / (n-3) = 6
==> multiply both sides by n-3
n = 6(n-3)
==> expand the right-hand side
n = 6n - 18
==> move like-terms to the same side
5n = 18
==> solve for n by dividing 18/5 to get our final solution
n = 3.6