Download MA 311: Exercises 1

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MA 311: Exercises 1
1. Assume that there are nine billiard balls, four of them are black and the
other five are colored blue, yellow, red, green, and orange. In how many
ways can one choose five balls out of these nine (so that with respect to the
black balls only their number is relevant)?
2. How many results are possible when throwing six identical 6-sided dice?
3. How many ways are there to roll two identical dice for a sum that is even?
4. What is the number of surjective functions from a 4-set to a 3-set? –
and if the selection is unordered, meaning that only the number of elements
reaching each member of the 3-set are relevant?
5. Prove the formula for any intergers n and k:
k
n =
k X
k
j=0
j
(n − 1)j .
How can this be used to prove by induction on n that the number of functions
from a k-set to an n set is nk ?
6. Prove the formula for n ≥ 1 and s ≥ 1:
s+n
n
=
s−1
s
s+1
s+n−1
+
+
+ ··· +
.
0
1
2
n
7. How many integers from 1 to 106 (inclusive) are either squares or cubes?
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MA 311: Answers to Exercises 1
P
5
= 1 + 5 + 10 +
1. Let k be the number of black balls. We get 4k=0 5−k
5
10 + 5 = 31 = 2 − 1, the number of subsets of a set of size 5 that exclude
the empty set. Easier is to notice that any choice of a non-empty subset of
the coloured balls can be augmented by the right number of black balls.
2. The throw of six identical dice yields six un-ordered choices of six different
values. Six objects
are divided by five dividers (for the six possible values),
for a total of 11
5 = 462.
3. It is easiest to look at each case.
To roll 2, there is only 1 + 1.
To roll 4, there are 1 + 3 and 2 + 2.
To roll 6, there are 1 + 5, 2 + 4 and 3 + 3.
To roll 8, there are 2 + 6, 3 + 5, and 4 + 4.
To roll 10, there are 4 + 6 and 5 + 5. To roll 12, only 6 + 6.
In total, there are 12 ways. But there are 21 ways to roll the dice, and yet
the probability of getting an even number is the same as getting an odd
number. The reason for this mismatch is simple: for each i ∈ {1, 2, 3, 4, 5, 6}
1
the probability to roll i + i is 36
while the probability to roll i + j for j 6= i
1
is 18 .
4.
formula (the littleStirlingformula) we have
P3Using our
i 3 (3 − i)4 = 34 − 3 24 + 3 = 81 − 16 + 3 = 36.
(−1)
i=0
i
1
2
If the names of the objects are not relevant, the formula gives 32 = 3. Three
positions must be filled, and then one of the three gets an extra object. This
is different from the number of paritions into three parts (where the positions
2
have no names), since that would be 36
3! = 6. Only the type [2 · 1 ] counts,
4
and there would be 2 = 6 choices for the partition member of size 2. (But
use the formula from Lemma 3.1 for more complicated types.)
5. Consider nk = ((n − 1) + 1)k and using the binomial expansion we get
k
n =
k X
k
j=0
j
j k−j
(n − 1) 1
=
k X
k
j=0
j
(n − 1)j .
Now prove that nk is the number of functions from a k-set to an n-set using
induction on n rather than on k. If |n| = 1 then there is exactly one function
and the formula is true. Let x be any fixed element of the n set and for
every such function f let x(f ) be the number of elements of the k set that
2
k
j
(n − 1)j different functions f such that
x(f ) = k − j, the (n − 1)j by the induction hypothesis and the kj being
the number of choices for k − j P
elements to be mapped to x. Therrefore by
induction hypothesis there are kj=0 kj (n − 1)j different functions from a
k-set to an n-set, and we have shown that this is equal to nk .
reach x. With j fixed there are
6. Order the possibilities of choosing a subset of size n from a set of size
s + n in the following way.
0: The first object is not chosen,
1: The first object is chosen, but not the second,
2: The first and second objects are chosen, but not the third, etc.
n: The first n objects were chosen.
This exhausts all the possibilities for choosing an n set
from an s + n set.
s+n−i−1
To any above category i there will be exactly
possibilities.
n−i
7. There are 1000 squares and 100 cubes. If a positive integer is both a
square and a cube then by unique factorization into primes it must be a
sixth power. With 10 numbers that are sixth powers there is a total of
1000+ 100-10= 1090.