Thanks for your comment. In the books question, order doesn’t matter: RN & NR count the same. For the denominator, the total number of pairs (45), I used a combination calculation — because order doesn’t matter. The 8 N-R pairs don’t count order — there would be 16 pairs if order did count. Suppose A, B, C, D are the novels, and X & Y are the reference works — the total set of N-R pairs, not counting order, is {AX, BX, CX, DX, AY, BY, CY, DY} — 8 pairs, and 8 more if order did count and we flipped all these around to get new orders . Does all this make sense?

Mike ]]>

Thanks for your wonderful post, I always read them to sharpen my probability skills.

However, I wish to clarify with you one concept in relation to above article w.r.t point 3),

I guess the no of way we can finally pick one reference and on e novel shouldn’t be => 4 * 2 / 2 = 4 because N R or R N is same selection isn’t it. If true, then should not the answer be -> 4/45? instead of 8/45?

Thanks once again for all the guidance, ]]>