solving for n yet again?

Question

solve for n, where n n is an element of real numbers
a) C(n, 2) = 45
b) (n+2)!/n! = 56

sorry n is an element of N so natural numbers not real sorry! :)

Answer 1

C(x, y) = x! / [y!(x-y)!]

so...
C(n, 2) = n! / [2!(n-2)!] = 45

Note that
n! = n * (n-1) * (n-2)!
and
2! = 2*1 = 2

So... we can simplify...
n! / [2!(n-2)!] = 45
[n*(n-1)*(n-2)!] / [2!(n-2)!] = 45
n(n-1) / 2 = 45
(n^2 - n) = 90
n^2 - n - 90 = 0
(n - 10)(n + 9) = 0
n = 10 or -9. We can eliminate -9, because we can't have a negative number in this case, so n = 10.

b)
(n+2)! = (n+2)(n+1)(n!)
so you can simplify the equation to be

(n + 2)(n + 1) = 56
n^2 + 3n + 2 = 56
n^2 + 3n - 54 = 0
(n + 9)(n - 6) = 0
n = -9 and 6

Again, we can eliminate -9, so n = 6

Answer 2

What do you mean by C (n, 2) - what does the C stand for?

About b):

(n+2)! = 1 * 2 * ... * n * (n+1) * (n+2)
n! = 1 * 2 * ... * n

So in the fraction nearly everything can be cancelled, and (n+1)(n+2) is left over:

(n+1)(n+2) = 56
n²+2n+n+2=56
n²+3n+2=56
n²+3n-54=0
(n+9)(n-6)=0

Thus, by Vietà's theorem the solutions are n=-9 and n=6

Edit: Ah, I just saw what you wrote. If n is a natural number, then the solution n=-9 is impossible. So n=6.

Answer 3

a) C(n,2) = 45
n(n-1)/2 = 45
n²-n = 90
n²-n-90=0
(n-10)(n+9)=0.
Since n is a natural number, n = 10.

b). This yields (n+1)(n+2) = 56
n²+3n -54 = 0
(n-6)(n+9) = 0.
Since n is a natural number, n = 6.

Answer 4

A.
C(n,2) = (n!)/[(n-2)!2!] which simplifies to (n)(n-1)/2.
So, if (n)(n-1)/2 = 45 then (n)(n-1) =90,
or,
n^2 - n - 90 = 0

--> n=10 or n = -9.

But, since C(n,2) is only defined for positive n, we have n=10.

B.
(n+2)!/n! simplifies to (n+1)(n+2). Following similar steps as part A we find that n =7 or n = -9 are possibilities, but reject the negative answer. Thus, n=7.

Answer 5

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

The first one can be rewritten as n!/(2!(n-2)!) = 45

Since 2! = 2 and n!/(n-2)! = n*(n-1) we can rewrite the problem as

n*(n-1)/2 = 45

n(n-1) = 90
n^2 - n - 90 = 0

Ignoring the negative root (because of the nature of the problem) n = 10


The second problem can be rewritten as (n+1)(n+2) = 56

n^2 + 3n + 2 = 56
n^2 +3n - 54 = 0

Ignoring the negative root, n = 6

Answer 6

a) n(n-1)/2!= 45
n^2-n-90= 0 n=((1+sqrt(361))/2 so n=10
b)(n+1)(n+2)=56

n^2+3n-54=0 n= ((-3+sqrt(225))/2= 6 n=6

Answer 7

a)

c(n,2) =45
n!/2!(n-2)!=45
n(n-1)(n-2)!/(n-2)!=45
n(squared)-n=2*45=90
solving this , we get,
n=10 , -9
since n belongs to natural numbers , hence we neglect -9 and accept 10 as the answer

b)
(n+2)!/n!=56
(n+2)(n+1)n!/n!=56
n(squared)+3n+2=56
solving this we get
n=6 ,-9
since n belongs to natural numbers , hence we neglect -9 and accept 6 as the answer.