Question 600383
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If 9 are both BB and UB, then of 12 UB, 3 of them are only UB.  Likewise of 14 BB, 5 of them are only BB.  In summary


37 neither, 9 both, 3 UB only, 5 BB only:  Total 54.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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