How many ways can we arrange $61$ numbered red cards and $61$ numbered blue cards so that cards of the same color cannot stand side by side?

Question

How many ways can we arrange $61$ numbered red cards and $61$ numbered blue cards so that cards of the same color cannot stand side by side?

I figured that you can place $61$ red cards in the first place, then you can place $61$ blue cards, then $60$ red cards... etc and then we also have that case when we start off with blue then red so the answer is:

$61!\times61!\times2$

Is this correct?

Answer 1

Yes! Think that you're basically just choosing an order on both the blue and red cards themselves, and then an order on which comes first in the row - blue or red.