permutation, combination and probability

Q.1 Different coloured pegs each of which is painted in one and only one of the six colours red, white,black,green, blue and yellow are to be placed in four holes with one peg in each hole.
Calculate how many different arrangements of pegs placed in the four holes so that they are all occupied can be made from 12 pegs, two of each colour.

my answer would be 12P4 / 2^6 but the answer in the book is 1170.

Q.2 The position of 9 trees which are to be planted along the sides of a road is five on the north side and four on the south side.
(a) find the number of ways in which this can be done if the trees are all of different species.

this was easy and I got the answer as 9!

(b) If the trees in (a) are planted at random, find the probability that two particular trees are next to each other on the same side of the road.

I tried solving this as 8! * 2! / 9! but the answer in the book seems to be 7/36.

(c) if there are 3 cupressus, 4 prunus and 2 magnolias, find the number of different ways in which these could be planted assuming that the trees of the same species are identical.

I got the answer to this one which is 1260 but I have a problem with next part.

(d) If the tress in (c) are planted at random, find the probability that the 2 magnolias are on the opposite sides of the road.

Reply 1

Dirac Delta Function

e-n-i-g-m-a

Break down the possibilities into different (mutually exclusive cases):

all different colours (1.1.1.1)
6C4.4! = 360

2 different colours and 2 the same (2.1.1)
6.5C2.4!/2! = 720

2 sets of the same colours (2.2)
6C2.4!/(2!2!) = 90

Sums to 1170

e-n-i-g-m-a

Q.2 The position of 9 trees which are to be planted along the sides of a road is five on the north side and four on the south side.
(a) find the number of ways in which this can be done if the trees are all of different species.

this was easy and I got the answer as 9!

(b) If the trees in (a) are planted at random, find the probability that two particular trees are next to each other on the same side of the road.

I tried solving this as 8! * 2! / 9! but the answer in the book seems to be 7/36.

That's because by lumping 2 of them as one and permuting across 8 positions, you are counting a split across the road as one position (think about them all arranged in a straight line rather than across a road).

So break it down across sides of the road: (4.7! + 3.7!)2/9! = 7/36

e-n-i-g-m-a

(c) if there are 3 cupressus, 4 prunus and 2 magnolias, find the number of different ways in which these could be planted assuming that the trees of the same species are identical.

I got the answer to this one which is 1260 but I have a problem with next part.

(d) If the tress in (c) are planted at random, find the probability that the 2 magnolias are on the opposite sides of the road.

The one in the north can be in 5 positions, while the south one can be in 4 positions, hence 20 permutations. each permutation is mulitplied by 7!/(3!4!) permutations of the other plants, hence 20.7!/(3!4!) = 700 out of 1260 permutations 700/1260 = 5/9