If by nCm you mean combinations of n things taken m at a time, you have a problem:
1C2 = 0
Assuming that is what you meant,
2n + 0 = [n(n-1)/2]^2
8n = n^4 - 2n^3 + n^2
n^4 - 2n^3 + n^2 - 8n = 0
Oops. no integer roots.
For the 2nd one,
12(n)(n-1)/2 = n!
6(n^2-n) = n!
Assume n>= 3
n^2 - n - n!/6 = 0
If n = 5, we have
n^2 - n - 20 = 0
(n-5)(n+4) = 0
so, n=5
12(5*4/2) = 120 = 5!
1) solve:
a) (2n+1C2) = (nC2)sqaured
b) 12(nC2) = n!
1 answer
6 answers