eaakbari wrote:
Bunuel wrote:
Practice: try to use the same concept.
Okay here goes,
The # of single letter codes possible would be \(n\) itself;
The # of pair of distinct letters codes possible would be (in alphabetical order); \(nC2\)
The # of Triples of distinct letters codes possible would be (in alphabetical order); \(nC3\)
Thus
\(nC3 + nC2 + n\)
> \(12\)
\(n*(n-1)/2 + n*(n-1)*(n-2)/3*2 + n\)
> \(12\)
Simplifying
\(n*(n^2 +5)\)
> \(72\)
Only sufficient value of \(n = 4\)
Is it correct?
Correct.
Three letters A, B, and C, are enough for 7<12 codes:
A;
B;
C;
AB;
AC;
BC;
ABC.
Four letters A, B, C, and D are enough for 15>12 codes:
A;
B;
C;
D;
AB;
AC;
AD;
BC;
BD;
CD;
ABC;
ABD;
ACD;
BCD;
ABCD.