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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 -

    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 -

    Thanks for your help! Could you please help me with the second problem because my answers are still not matching?

  • Maths -

    "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

  • Maths -

    Thanks a lot for your help!!

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