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IMO C
suppose 1) Language Q could have both 3 or 2 code symbols, for example. The code group would be in the two cases:
a) A,B, AB;
b) A,B,C,AB,AC,BA,BC,CA,CB,ABC,ACB,BAC,BCA,CAB,CBA;
It is consisten with 1
so INSUFFICIENT
suppose 2) could, again, be both the language of all of the horizontal arrangements of one or more distinct code symbol among A,B,C (b) and the language of one code symbol among 15 code symbols (A;B;C;D;E;F;G;H;I;J;K;L;M;N,O)-> INSUFFICIENT
suppose 1) and 2) the number of code groups with these assumptions are f(n)=n+n*(n-1)+n*(n-1)*(n-2)+...+n! where n is the number of code symbols.
For example for n=3 we have 3 single code symbols (A,B,C), 3*2 2-code symbols strings (AB,AC,BA,BC,CA,CB) and 3*2*1
3-code symbols strings (ABC,ACB,BAC,BCA,CAB,CBA)
3+3*2+3*2*1=15 so 3 code symbols fit well our problem. However f(n) in increasing so only n=3 can be the answer so this is SUFFICIENT.

Ah ok I didn't read that the official answer was C

As is clear that Statement 1 & 2 cannot answer independently. So lets combine.

Assume there are only 3 code symbols and we wish to make distinct code groups using distinct symbol only.

Code group with 1 symbol only = 3.
Code group with 2 symbol only = 3P2 = 3.2 = 6.
Code group with 3 symbol only = 3P3 = 3.2.1 = 6.

Totaling 15. It satisfies both the statements hence ans C.

_________________
Shalabh Jain,
e-GMAT Instructor