How many numbers of 6 digits which have exatctly the digit 1 (2 times), digit 2 (2 times), without zero, are there?(adsbygoogle = window.adsbygoogle || []).push({});

The book post this solution: [tex] \frac{6!}{2!2!}*\binom{7}{2} + \frac{6!}{2!2!2!}*7= 4410[/tex],

but i'm trying to find an explanation for this result.

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# Combinatorial question: permutation, binomial coefficient

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