Question 1170989
<pre>
Hey, tutors, you two should be ashamed of yourselves! Leave this student alone!
Those parentheses he or she had up there meant combinations, or if you prefer,
binomial coefficients.  I rewrote it up there like he/she obviously meant it. 

{{{(matrix(2,1,N,K))}}} means the same as C(N,K) or NCK. 

The simplest way to evaluate combinations, or binomial coefficients, is not by
using the definition, which is: 

{{{(matrix(2,1,N,K))= N!/(K!(N-K)!)}}} 

but by this:

{{{(matrix(2,1,N,K))= (N*(N-1)*""*""*(N-K+1))/(K*(K-1)*""*""1)}}}, where there are the same number of factors in the numerator 
as in the factorial (K!) in the denominator.

So

{{{matrix(1,3,

4*(matrix(2,1,n,2)),
""="",
(matrix(2,1,n+2,3)))}}}

{{{matrix(1,3,

4*expr((n(n-1))/(2*1)),
""="",
((n+2)(n+1)(n))/(3*2*1)))}}}

{{{matrix(1,3,

2*n(n-1),
""="",
((n+2)(n+1)(n))/(6)))}}}

{{{matrix(1,3,

12*n(n-1),
""="",
(n+2)(n+1)n))}}}

Divide both sides by n:

{{{matrix(1,3,

12*(n-1),
""="",
(n+2)(n+1)))}}}

{{{matrix(1,3,

12n-12,
""="",
n^2+3n+2))}}}


{{{matrix(1,3,

0,
""="",
n^2-9n+14))}}}

{{{matrix(1,3,

0,
""="",
(n-7)(n-2))}}}

n-7=0; n-2=0
  n=7;   n=2

So there are two solutions.

They both check:


{{{matrix(1,3,

4*(matrix(2,1,7,2)),
""="",
(matrix(2,1,7+2,3)))}}}

{{{matrix(1,3,

4*(matrix(2,1,7,2)),
""="",
(matrix(2,1,9,3)))}}}

{{{matrix(1,3,

4*21,
""="",
84))}}}

-------------

{{{matrix(1,3,

4*(matrix(2,1,2,2)),
""="",
(matrix(2,1,2+2,3)))}}}

{{{matrix(1,3,

4*(matrix(2,1,2,2)),
""="",
(matrix(2,1,4,3)))}}}

{{{matrix(1,3,

4*1,
""="",
4))}}}

Edwin</pre>