solving for n when...?

Question

a) n!/(n-2)! = 930 ~~ I got n= 930 for this ??
b) P(n,5) = 42x P(n,3) ~~ i have no idea where to even start here

Answer 1

a) You can cancel (n-2)! from both. To demonstrate:



n! = n * (n-1) * (n-2) * (n-3) * .... 2 * 1

(n-2)! = (n-2) * (n-3) * .... 2 * 1



So, you can cancel out (n-2)! to get n * (n-1). You now have:

n(n-1) = 930

n^2 - n - 930 = 0

(n - 31)(n + 30) = 0

So, n = -30 or n = 31.



Since you can't have factorials of negative numbers, n must equal 31.



b) Write out the Permutations.

P(n, 5) = n!/(n-5)!

42 * P(n, 3) = 42n!/(n-3)!



Set them equal to each other.



n!/(n-5)! = 42n!/(n-3)!



You can cancel the n! to get:

1/(n-5)! = 42/(n-3)!



Examine what (n-3)! means:

(n-3)! = (n-3) * (n-4) * (n-5) * (n-6) *... * 2 * 1

You can see (n-5)! in there, so you know you can cancel (n-5)! to get:

1/1 = 42/(n-3)(n-4)



Get everything on one side to create a quadratic equation:

(n-3)(n-4) = 42

n^2 - 7n +12 = 42

n^2 - 7n - 30 = 0

(n - 10)(n + 3) = 0

So, n = 10 or n = -3.



Once again, we can't have negative solutions, so n must equal 10.

Answer 2

n!/(n-2)!=930



=>n(n-1)(n-2).../((n-2)(n-3)(n-4)..=930

=>n(n-1)=n^2-n=930

=>n^2-n-930=0



Quadratic solution!



n=(1 +/- sqrt(1+4*930))/2=1/2 +/- 1/2sqrt(3721)

n=1/2 +/- 61/2

n=31 or -30



choose the positive n=31, since factorials are not defined for negative numbers (not at this level anyway!)



The second part of your question's too vague. What's the form of the probability function P(n,m)?

Answer 3

a) n!/(n-2)! = n*(n-1) = 930

n^2 - n - 930 = 0

(n-31)(n+30) = 0

n= 31

(Ignore the n=-30 answer because the factorial function is only defined for whole numbers)



b) n!/[(n-5)!] = 42*n!/[(n-3)!]

n*(n-1)*(n-2)*(n-3)*(n-4) = 42*n*(n-1)*(n-2)/

(n-3)(n-4) = 42

n^2 - 7n + 12 = 42

n^2 - 7n - 30 = 0

(n-10)(n+3) = 0

n = 10

(Ignore n = -3)

Answer 4

a) That would be for n!/(n-1)!...



n!/(n-2)! = n * (n-1) = 930



n = 31 (easy shortcut -- take the square root and round up, then check the answer)



b) P(n, 5) means n!/(n-5)!. n! is n * n-1 * n-2 * ... Everything from n-5 on down will be cancelled by the (n-5)!, leaving you with just: n * (n-1) * (n-2) * (n-3) * (n-4)



Similarly, P(n, 3) leaves you with the first three terms. Thus:

(n-3) * (n-4) = 42

and...

n = 10

Answer 5

n(n-1)(n-2)!/(n-2)! = 930

cancel out (n-2)!

you got

n(n-1)=930

n^2-n-930=0

(n+30)(n-31)=0

n= -30 not accepted

n=31 accepted

Answer 6

i have no idea how to do that, but are u taking chemistry throuigh correspondence? because some of those questions look familiar. if you are could u email me or something and we could help eachother out! thanks so much!!, LaLa