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MTH6109
Combinatorics
Solutions 1
1
(i) Each element of the sequence can be chosen in 7 ways, independently of
the other elements of the sequence. Hence by the AND Principle, there are
75 = 16807 sequences.
(ii) The first element can be chosen in 7 ways, the next in 6, the third in 5, the
fourth in 4 and the fifth in 3, so the number of sequences is 7.6.5.4.3 = 2520.
(iii) We can choose the first odd number in 4 ways, and the second in 3. Then
we choose the first even number in 3 ways, the second in 2, and the third
in 1. Hence by the AND Principle, the answer is 4.3.3.2.1 = 72.
(iv) There are two sorts of such sequences, the ones which start with an odd
number, so alternate odd-even-odd-even-odd, and those which start with an
even number, so alternate even-odd-even-odd-even. The number of the first
type is 4.3.3.2.2 = 144, and the number of the second type is 3.4.2.3.1 = 72,
so the total number is 144 + 72 = 216, by the OR Principle.
(v) There are various ways to answer this question. My suggestion is to first
choose the set of five numbers in the sequence, by first choosing
two
from
3
4
{1, 2, 3}, and then choosing three from {4, 5, 6, 7}, making 2 . 3 = 12
choices so far; and then order the five numbers in 5! = 120 ways, so that
the final answer is 12.120 = 1440.
2
(i) Since X has size 9, the total number of subsets of size 5 is
9.8.7.6/4.3.2.1 = 126.
(ii) We can choose the two vowels in
in
6
3
3
2
9
5
=
9
4
=
= 3 ways, and the three consonants
= 20 ways, making 2.30 = 60 choices in all.
1
(ii) As well as the sets with two vowels, which we have just counted, we have
the sets with all three vowels.
The number of these is the number of choices
of two consonants, that is 62 = 15. Hence by the OR Principle the total
number is 60 + 15 = 75.
(iii) There are 10 letters, of which I and L are repeated 3 times each. Hence
there are 10!/3!3! = 100800.
3 If n is odd there is an easy way to do this, by mapping every even subset to
its complement, which is odd.
In general, there is a trick which involves taking a fixed element from A, say
1 ∈ A, and mapping even sets B to odd sets f (B) by defining f (B) = B ∪ {1} if
1 6∈ B, and f (B) = B \ {1} if 1 ∈ B. In other words, we either add in 1 or take
it out according to whether it is not in or in B.
Either way, we see that the number of even subsets is equal to the number of
odd subsets. Since the total number of subsets is 2n , the number of odd subsets
is 2n−1 .
There is another method of proving this result using the binomial theorem to
show that (1 − 1)n is equal to the number of even subsets minus the number of
odd subsets. However, this does not answer the question as posed.
2