The correct option is A (10ab)12
Need to simplify:–––––––––––––––––––––163a12×1254b12
163a12×1254b12
=(24)3a12×(53)4b12
[ As 16=2×2×2×2=24 and 125=5×5×5=53 ]
=24×3a12×53×4b12
[ As (am)n=am×n ]
=212a12×512b12
=212×a12×512×b12
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[ Multiplying exponents with different bases but same exponent: cn×dn×pn×qn=(c×d×p×q)n ]
=(2×a×5×b)12
=(2×5×a×b)12
⇒163a12×1254b12=(10ab)12
Therefore, option (a.) is the correct one.