We know the equation carry look ahead generator circuit is
$c_{i}=g_{i-1}+c_{i-1}.p_{i-1}$
$c_{1}=g_{0}+c_{0}.p_{0}$
$c_{2}=g_{1}+c_{1}.p_{1}$
$\Rightarrow c_{2}=g_{1}+\left ( g_{0}+c_{0}.p_{0}\right )p_{1}$
$\Rightarrow c_{2}=g_{1}+g_{0}.p_{1}+c_{0}.p_{0} .p_{1}$
$c_{3}=g_{2}+c_{2}.p_{2}$
$\Rightarrow c_{3}=g_{2}+\left (g_{1}+g_{0}.p_{1}+c_{0}.p_{0} .p_{1}\right )p_{2}$
$\Rightarrow c_{3}=g_{2}+g_{1}.p_{2}+g_{0}.p_{1}.p_{2}+c_{0}.p_{0} .p_{1}.p_{2}$
$c_{4}=g_{3}+c_{3}.p_{3}$
$\Rightarrow c_{4}=g_{3}+\left (g_{2}+g_{1}.p_{2}+g_{0}.p_{1}.p_{2}+c_{0}.p_{0} .p_{1}.p_{2}\right )p_{3}$
$\Rightarrow c_{4}=g_{3}+g_{2}.p_{3}+g_{1}.p_{2}.p_{3}+g_{0}.p_{1}.p_{2}.p_{3}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}$
$c_{5}=g_{4}+c_{4}.p_{4}$
$\Rightarrow c_{5}=g_{4}+\left (g_{3}+g_{2}.p_{3}+g_{1}.p_{2}.p_{3}+g_{0}.p_{1}.p_{2}.p_{3}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}\right )p_{4}$
$\Rightarrow c_{5}=g_{4}+g_{3}.p_{4}+g_{2}.p_{3}.p_{4}+g_{1}.p_{2}.p_{3}.p_{4}+g_{0}.p_{1}.p_{2}.p_{3}.p_{4}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}p_{4}$
Total 15 AND gates.
if total AND gates were asked in whole carry-look ahead adder
, than it would be n*(n+1)/2 + n.
But here it is asking, The number of AND gates are present inside a 5-bit carry look ahead generator circuit
are $15$