posted by Claire on .
These questions are about permutations and combination:
Find the number of ways in which a class of 20 children can be divided into two groups of 6 and 14 if the class has a set of twins who are inseperable. (ans. is 21624)
Four books are taken from a shelf of 18 of which 6 are paperback and 12 are hardback. In how many ways can 4 books be chosen if at least one of them is paperback? (ans, is 2565)
Can somebody help me to work these questions out?
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!!