Posted by **Claire** on Wednesday, October 8, 2008 at 12:40pm.

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?

- Maths -
**Reiny**, Wednesday, October 8, 2008 at 1:33pm
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

- Maths -
**Reiny**, Wednesday, October 8, 2008 at 2:10pm
"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

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