In NaCl like structure B covers all the corners and face centre and A covers all the edge centres and body centre

Net contribution of B = 8*1/8 + 6*1/2 = 4

Net contribution of A = 12*1/4 + 1*1 = 4

now 2 A ATOMS ARE REMOVED FROM EDGES so remaining contribution = 10*1/4 + 1*1 = 7/2

and 2 B ATOMS ARE REMOVED FROM CORNE so remainnig contribution = 6*1/8 + 6*1/2 = 15/4

so formula will be A_7/2B_15/4  or A₁₄B₁₅