Indian Statistical Institute B.Math & B.Stat : Combinatorics

Among all the factors $4^6 6^7 21^8$ the number of factors which are perfect squares is? $4^6 6^7 21^8 = 2^{12} \times 2^7 \times 3^7 \times 3^8 \times 7^8 = 2^{19} \times 3^{15} \times 7^8$. Now a factor of $4^6 6^7 21^8$ must be of the form $2^i3^k7^k$ where $i,j,k$ are integers and $0 \leq i \leq 19, 0 \leq j \leq 15 \quad and \quad 0 \leq k \leq 8$, since the problem asks for divisors which are perfect squares $i,j,k$ must be even. Now conider the product $(2^0+ 2^2+ \dots + 2^{18})(3^0+ 3^2+ \dots + 3^{14})(7^0+ 7^2+ \dots + 7^8)$. Each term of the product satisfies the above two condtion, so the required number of factors which are perfect squares is equal to the number of terms of the above product, which in turn equals to $10 \times 8 \times 5 = 400$