I'll do the first one, you do the second one the same way.
the twins are either in the small group or the large group
case1: twins in small group
that leaves 4 other to choose from the remaining 18, the rest will make up the larger group
---> C(18,4) = 3060
Case2: twins are in the larger group
that leaves 12 to be chosen from the remaining 18, the rest will make up the smaller group
----> C(18,12) = 18564
for a total of 3060+18564 = 21624
Thanks for your help! Could you please help me with the second problem because my answers are still not matching?
"at least one paperback (pb)"
means 1 pb, or 2 pb, or 3 pb, or 4 pb.
so direct way:
1 pb, 3hb ---> C(6,1) x C(12,3) = 1320
2 pb, 2 hb ---> C(6,2) x C(12,2) = 990
3 pb, 1 hb ---> C(6,3) x C(12,1) = 240
4 pb, 0 hb ---> C(6,4) x C(12,0) = 15
total = 2565
A shorter way would be doing it the "back door way":
with no restrictions ---> C(18,4) = 3060
the case we DON'T want is
0 pb, 4 hb = (6,0)xC(12,4) = 495
no number of ways with at least one pb is 3060 - 495 = 2565
Thanks a lot for your help!!
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