Download Combinatorics

Combinatorics
Homework A
Name: Sam Poole
January 16, 2007
A7
Find the number of sequences of length 4 in which the first two terms are letters from the
alphabet {A, B, C, …, Z} and the last two terms are digits from the set
{0, 1, 2, …, 9} with the condition that the two digits must be different.
Solution. We can imagine filling in four blanks ___ ___ ___ ___, left to right. here are 26
choices for the first blank, 26 choices for the second blank, 10 choices for the third blank and 9
choices for the fourth blank. By the Product Rule, there are (26)(26)(10)(9) such sequences.
A19 Suppose the numbers 1, 2, 3, 4, 5 are arranged in a random order. What is the probability
that all odd numbers still appear in odd numbered positions.
Solution. We can imagine filling in five blanks ___ ___ ___ ___ ___, left to right, to create the
(equally likely) outcomes. The total number of these is 5! = 120, by the Product Rule. To count
the number of these satisfying the given condition, simply adapt the Product Rule accordingly:
Filling in left-to-right, there are 3 choices for position one, 2 choices for position two, 2 choices
for position three, 1 choice for position four, and then 1 choice for position five. Hence, the
number of outcomes satisfying the given criterion is (3)(2)(2)(1)(1) = 12, which makes the
12
1

probability of this happening
.
120 10
A20c Find all derangments of ABCD.
Solution. There aren’t that many, so we will just list all 9 of them:
BADC
DABC
BCDA
DCAB
BDAC
DCBA
CADB
CDBA
CDAB
A21 Find the number of rearrangements of ABCDE having exactly one letter in its original
position.
Solution. Suppose the letter A is the letter in its original position. Hence, the arrangement looks
like
A____
The number of ways to fill in the remaining blanks so that no other letter is in its original
position is 9 (the value of D4) according to the solution to problem A20 above. Similarly, if the
letter that remains in its original position is B, then there are 9 ways to complete the other four
blanks. This is true for any of the five letters that remain in their original position. Therefore,
the total number of arrangements is 5 (the number of choices for which letter stays fixed) times 9
(the number of derangements of the other four letters). This gives a final answer of 45.