Arrangement Problem Solutions: 10 Trees

[coldcell]

1. Ten trees - four pines, four cedars, and two spruce- are planted in two parallel rows of five trees. How many arrangements are possible if each row must have the same composition of trees, not necessarily in the same order.

2. Suppose we want to creat subsets of the ten digits {0, 1, 2, ..., 9}
a) How many subsets can be created, including the empty set?
b) How many of the subsets contain only digits less than 7?
c) How many of the subsets contain 0 or 9?

 

[HallsofIvy]

Since you are posting in this section, presumably you have read the guide-lines. What have you done on this problem yourself?

Oh, and is this really from a "calculus and beyond" course?

 

[coldcell]

Just read the guide-lines :S. Actually I'm in Grade 12 and since I took Calculus last semester, I thought these questions are appropriate in this forum. Where should I have put it?

1. I tried many things, but the closest answer I can get is 900. Since the trees of each kind are the same, we only need to concern ourself with the first row, then simply square it.

So, I use 5!/(2!2!). 5 spots, 2 alike, 2 alike. This gives me 30. This times the second row, since for every combination of the first row, there is another combination of the second row.

Hence 30² = 900

However the answer at the back of the book is 225. No matter how I try, I simply couldn't get the right answer.

2. a) I solved this but I used a long method. I simply do (10 C 0) + (10 C 1) + (10 C 2) +...+ (10 C 10). This gives me the proper answer. I'm wondering if there is a shorter way of doing this.

b) and c), I simply have no idea.

 

[HallsofIvy]

My point was that you must show what you have tried first!

I get exactly the same as you on the first problem: 900.
Is it possible that they don't want to count arrangements which have the same trees in the same order, but with the rows swapped, as different?
That is: is
PCPCS
CPCPS considered the same as

CPCPS
PCPCS and so not counted separately?

For the second problem, the only question is "for each digit is it included in the subset or not?" That is, there are 10 digits for each of which you have 2 choices: how many choices is that? (You do recognize, don't you, that nC0+ nC1+ ...+ nCn= 2ⁿ? That's always true because it's just the sum of the binomial coefficients: (x+ y)ⁿ with x and y both equal to 1.)

Do the same for (b) and (c). In (b), the only digits to be considered are {1, 2, 3, 4, 5, 6} and for (c) only {0, 9}.