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Cherokee
science forum beginner
Joined: 12 Jun 2006
Posts: 1
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 Posted: Mon Jun 12, 2006 6:45 pm Post subject:
Quick Probability/Counting Technique Questions
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Hey there, I've got a few questions that are giving me trouble.
1. How many even numbers greater than 4000 can be formed using some or all of the digits 1,2,3,4,5,6 if each digit must feature no more than once in a number?
2. Security codes are to be allocated with each code being either 2 digits followed by 3 letters OR 3 digits followed by 2 letters.
eg. 12AQT or 129PT
Any of the digits 0 to 9 can be used and any of the letters A-Z can be used but no digit or letter may be repeated.
Determine how many codes are possible in each of the following cases
e)A code must contain a 9
f)A code must contain a 5
g)A code must contain at least one of the digits 5 and 9
-These problems relate to permutations and combinations.. Counting Techniques.
3. A child is told she can bring 5 toys with her on a holiday. The child decides to choose the 5 from
6 jigsaws
8 dolls
4 balls
2 trucks
How many of these sets have at least one from each of the four categories listed above?
Any at all help is much appreciated,
Many thanks in advance. |
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Brian M. Scott
science forum Guru
Joined: 10 May 2005
Posts: 332
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| Posted: Mon Jun 12, 2006 8:35 pm Post subject:
Re: Quick Probability/Counting Technique Questions
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On Mon, 12 Jun 2006 14:45:11 EDT, Cherokee
<wartex11@iinet.net.au> wrote in
<news:32595816.1150137941100.JavaMail.jakarta@nitrogen.mathforum.org>
in alt.math.undergrad:
| Quote: | Hey there, I've got a few questions that are giving me trouble.
|
In general it's best to give some indication of what you've
already tried; quite apart from the fact that people like to
know that you *have* made a serious effort, it can save you
from being told a lot of things that you already know.
None of these requires any kind of tricky technique or
analysis. In each you have a set of things to be counted,
and the problem can be solved by some combination of three
standard techniques:
* breaking up that set into non-overlapping smaller
sets that can be counted more easily and then
adding the subtotals;
* counting some bigger set, counting the number of
things in that bigger set that *aren't* in the one that
you're trying to count, and taking the difference; or
* counting the things in each of two overlapping
subsets, adding these partial results, and then
subtracting from this preliminary total the number
of things that are in the overlap and so have been
counted twice.
| Quote: | 1. How many even numbers greater than 4000 can be formed
using some or all of the digits 1,2,3,4,5,6 if each digit
must feature no more than once in a number?
|
Every such number will obviously have four, five, or six
digits. Imagine choosing one of them one digit at a time,
starting at the righthand end. Since the number is to be
even, it must end in 2, 4, or 6, so there are 3 ways to
choose the last digit. Once that's chosen, there are 5
digits still available, and you can use any of them to fill
the 2nd position from the right; altogether that's 5*3 = 15
ways to fill the rightmost two slots. You've now used two
of the six available digits, so you have only 4 possible
choices for the 3rd slot; this means that there are 4*5*3 =
60 ways to fill the rightmost three slots.
At this point it's easy to go astray by trying to do too
much at once. Don't try to take care of the 'greater than
4000' requirement at this point; for now just count all of
the even numbers having four or more digits. We've used up
three of the six available digits, so there are 3 ways to
fill the 4th slot, and hence 3*4*5*3 = 180 four-digit even
numbers that use only the digits 1 through 6 and use each
digit at most once. If you continue in this fashion, you
can easily calculate the number of five-digit and six-digit
numbers of this type; all of them are clearly greater than
4000 and should therefore be included in your final total.
However, the 180 four-digit numbers that we counted include
some that are less than 4000; these have to be excluded from
the final total, so we need to count them.
Clearly they all begin with 1, 2, or 3, and they all end
with 2, 4, or 6. If one of them begins with 1 or 3, it can
end in any of 2, 4, and 6; if it begins with 2, however, it
can end only in 4 or 6, so it's simplest to split the
counting into two cases. If the number begins with 1 or 3,
there are 2 ways to fill the leftmost slot and 3 ways (2, 4,
or 6) to fill the rightmost slot. That leaves four digits,
so there are 4 possible choices for the 2nd slot from the
left and 3 possible choices for the remaining slot, for a
total of 2*3*4*3 = 72 four-digit even numbers beginning with
1 or 3, using only the digits 1 through 6, and using each of
them at most once. In similar fashion you can calculate the
number that begin with a 2, but I'll let you do that.
| Quote: | 2. Security codes are to be allocated with each code being
either 2 digits followed by 3 letters OR 3 digits
followed by 2 letters. eg. 12AQT or 129PT
Any of the digits 0 to 9 can be used and any of the
letters A-Z can be used but no digit or letter may be
repeated.
Determine how many codes are possible in each of the
following cases
e)A code must contain a 9
|
Break the count into two parts, just as you would if there
were no such restriction. How many codes are there with
three digits and two letters? If we were allowed to use a
9, there would be 10*9*8 = 720 possible combinations of
digits. How many of them include a 9? There are 9*8 = 72
three-digit strings of the form 9xy (with x not equal to y).
There are 72 more of the form x9y and another 72 of the form
xy9. No three-digit string is in more than one of these
categories, so there are 3*72 = 216 'bad' strings, and hence
720 - 216 = 504 usable three-digit combinations. Each can
be combined with any pair of distinct letters, for a total
of 504*26*25 codes with three digits and two letters. The
calculation of the number of codes of the other type is
similar. The two types don't overlap, so you can simply add
the two partial results to get the desired answer.
| Quote: | f)A code must contain a 5
|
This is of course the same kind of problem as (e). In fact,
if you think about it, you should be able to see why it's
essentially the *same* problem, not just the same kind, and
why it has the same answer.
| Quote: | g)A code must contain at least one of the digits 5 and 9
|
Start by adding your answers to (e) and (f). Of course this
doesn't quite work, because every code that contains both a
5 and a 9 gets counted twice. To correct this double
counting, you need to subtract from that total the number of
codes containing both a 5 and a 9; this number can be
calculated using exactly the same techniques that are used
to answer (e) and (f).
| Quote: | -These problems relate to permutations and combinations..
Counting Techniques.
3. A child is told she can bring 5 toys with her on a
holiday. The child decides to choose the 5 from
6 jigsaws
8 dolls
4 balls
2 trucks
How many of these sets have at least one from each of the
four categories listed above?
|
The problem statement is ambiguous: the answer depends on
whether the items of each type are identical (so that the
four balls, for instance, are completely interchangeable) or
distinguishable. I'm going to assume that they're
distinguishable.
Clearly she must take two of one kind of toy and one of each
of the others. The most straightforward solution, if not
the quickest, is to divide the problem into four cases
according to the type of toy of which she takes two.
Two jigsaws: There are C(6, 2) = 6!/(2!*4!) = 15 ways to
choose 2 of the 6 jigsaws. There are then 8 ways to choose
a doll, 4 ways to choose a ball, and 2 ways to choose a
truck, for a grand total of 15*8*4*2 possible selections.
Do the other cases similarly and then add them up.
Brian |
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