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The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n with parts modulo 2,3,6,9,10 modulo 12

This is from Subbarao 1971 but I don't quite understand the method, or I have possibly misunderstood what the answer is supposed to be?

So I have $$\prod^\infty_{n=1}\,\,(1 + q^{2n} + q^{3n} + q^{5n})$$

I believe this should eventually be $$\prod^\infty_{n=1} \frac{1}{(1-q^{12n+2})(1-q^{12n+3})(1-q^{12n+6})(1-q^{12n+9})(1-q^{12n+10})}$$ But I'm finding it impossible to get from A to B. I'm pretty new to this stuff.. Can anybody help?

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The trick is the same as in the proof that the numbers of partitions with odd parts and with distinct parts are the same. Then $$ \begin{split} \prod_{n=1}^{\infty}\left(1+q^{2n}+q^{3n}+q^{5n}\right) &=\prod_{n=1}^{\infty}\left(1+q^{2n}\right)\left(1+q^{3n}\right)=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{1-q^{2n}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{\left(1-q^{4n}\right)\left(1-q^{4n+2}\right)}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{1-q^{4n+2}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+10}\right)}\frac{\left(1-q^{12n}\right)\left(1-q^{12n+6}\right)}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+10}\right)}\frac{1-q^{12n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+10}\right)}\frac{1}{\left(1-q^{12n+3}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+9}\right)}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+3}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+9}\right)\left(1-q^{12n+10}\right)}, \end{split} $$ as desired.

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