The following question marks a student answer as incorrect in part (d) unless the factorial factors are typed first. For example, the answer for seed 1282 is 25*26!*4!, which is marked as incorrect unless entered as 26!*4!*25.

I cannot figure out why. Any suggestions welcome.

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## DBsubject(Discrete Mathematics)

## DBchapter(Permutations and Combinations)

## DBsection(Permutations)

## Level()

## KEYWORDS(permutations)

## TitleText1(Discrete Mathematics With Graph Theory)

## EditionText1(3)

## AuthorText1(Goodaire and Parmenter)

## Section1(7.1)

## Problem1(7)

## Author(Tim Alderson)

## Institution(University of New Brunswick Saint John)

## Language(en-CA)

DOCUMENT();

####################

# Load Macros

####################

loadMacros(

"PGstandard.pl",

"MathObjects.pl",

"PGML.pl",

"AnswerFormatHelp.pl",

"PCCmacros.pl",

"contextIntegerFunctions.pl",

"PGcourse.pl",

);

####################

# Header

####################

TEXT(beginproblem());

####################

# PG Setup

####################

Context("IntegerFunctions");

Context()->flags->set(tolerance => 0.1, tolType => "absolute");

$b = random(8,26); #define number of possible elements in set S

do {$g = random(2,8);} until ($g != $b);

$n = Compute($b+$g); #

$nm1 = $n-1; #

$np1 = $n+1; #P($bp1,n)

$bm1 = $b-1; #

$bp1 = $b+1; #

$gm1 = $g-1; #

$gp1 = $g+1; #

$ans1 = Compute("[$n]!");

$ans2 = Compute(" [$b]!*[$g]!*2!");

$ans3 = Compute("[$g]!*[$bp1]!");

$ans4 = Compute("([$g]!)*([$b]!)*([$bm1])") ;

####################

# Body

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BEGIN_PGML

In how many ways can [$b] boys and [$g] girls sit in a row:

a. with no restrictions?

[_____]{$ans1}

a. If the boys sit together and the girls sit together.?

[_____]{$ans2}

a. If the girls sit together.?

[_____]{$ans3}

a. If just the girls sit together.?

[_____]{$ans4}

END_PGML

####################

# Solution

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BEGIN_PGML_SOLUTION

a. This question asks for the number of ways to arrange [$n] people in a row.

The answer is [$n]!.

a. Step 1: Arrange the boys in a line: [$b]! ways

Step 2: Arrange the girls in a line: [$g]! ways

Step 3: Put either the boys or the girls at the start of the line: 2 ways.

The answer is 2([$b]!)([$g]!).

a. Step 1: Arrange the girls in one of [$g]! ways.

Step 2: The girls form one block and the boys ten other blocks.

These [$bp1] blocks can be arranged in one of [$bp1]! ways.

The answer is [$g]![$bp1]!.

a. Let [`A`] be the set of ways in which the girls sit together and [`B`] the set of ways

in which the boys and girls each sit together.

The question asks for [`|A\setminus B|`]:

[``|A\setminus B|=|A| - |A\cap B|``]

[`|A|=[$g]![$bp1]!`] (from part (c)), and [`|A\cap B|=2([$b]!)([$g]!)`] (from part (b)).

Therefore the answer is [$g]![$bp1]!-2([$b]!)([$g]!)= [$bm1]([$g]![$b]!).

END_PGML_SOLUTION

####################

# End Problem

####################

ENDDOCUMENT();