{
"cells": [
 {
		   "cell_type": "markdown",
	   "metadata": {},
	   "source": [
       "# Chapter 6: Counting"
	   ]
	},
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.14: Ordered_partitions.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"c1=3; //number of toys that the youngest child should get\n",
"c2=2;  //number of toys that  the third child should get\n",
"c3=2;  //number of toys that the second child should get\n",
"c4=2;   //number of toys that the eldest son should get\n",
"m=factorial(9)/(factorial(3)*factorial(2)*factorial(2)*factorial(2));\n",
"disp(m,'number of ways nine toys can be divided between four children with the youngest son getting 3 toys and others getting 2 each')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.15: Unordered_partitions.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"p=12;  //total number of students\n",
"t=3;  //number of teams or partition\n",
"disp('each partition of the students can be arranged in 3! ways as an ordered partition')\n",
"r=factorial(12)/(factorial(4)*factorial(4)*factorial(4))  //number of ordered partitions\n",
"m=r/factorial(t);  //number of unordered partitions\n",
"disp(m,'number of ways that 12 students can be partitioned into three teams so that each team consists of 4 students')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.16: Inclusion_exclusion_principle_revisited.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"U=1000;  //number of elements in the set of positive integers not exceeding 1000\n",
"A=U/3;   //number of elements in the subset of integers divisible by 3\n",
"B=U/5;   //number of elements in the subset of integers divisible by 5\n",
"C=U/7;   //number of elements in the subset of integers divisible by 7\n",
"AandB=floor(U/(3*5))   //number of elements in the subset containing numbers divisible by both 3 and 5\n",
"AandC=floor(U/(3*7))   //number of elements in the subset containing numbers divisible by both 3 and 7\n",
"BandC=floor(U/(5*7))   //number of elements in the subset containing numbers divisible by both 5 and 7\n",
"AandBandC=floor(U/(3*5*7))  //number of elements in the subset containing numbers divisible by 3,5 and 7\n",
"s=U-(A+B+C)+(AandB+AandC+BandC)-(AandBandC);  // By inclusion-exclusion principle\n",
"S=round(s);\n",
"disp(S,'The number of integers in the set U, which are not divisible by 3,5 and 7 is')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.1: Sum_rule_principle.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"M=8;    //number of male professors teaching calculus\n",
"F=5;    //number of female professors teaching calculus\n",
"T=M+F ;\n",
"disp(T,'number of ways a student can choose a calculus professor')\n",
"\n",
"E=[2,3,5,7];   //event of choosing a prime number less than 10\n",
"F=[2,4,6,8];   //event of choosing an even number less than 10\n",
"G=intersect(E,F);  //event of getting an even and prime number \n",
"H=length(E)+length(F)-length(G);  \n",
"disp(H,'event of getting an even or a prime number')\n",
"\n",
"E=[11,13,17,19]; //event of choosing a prime number between 10 and 20\n",
"F=[12,14,16,18];  //event of choosing an even number between 10 and 20\n",
"G=union(E,F);    //event of choosing a number which is prime or even\n",
"k=length(G); \n",
"disp(k,'number of ways of choosing a number which is prime or even')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.2: Product_rule_principle.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"disp('a license plate contains two letters followed by three digits where first digit can not be zero') \n",
"n=26; //number of english letters\n",
"n*n;  //number of ways of choosing two letters in the license plate\n",
"p=10; //number of digits (0-9)\n",
"(p-1)*p*p; //number of ways to select the three digits with the first digit not being zero\n",
"k=n*n*(p-1)*p*p;\n",
"disp(k,'total number of license plates that can be printed')\n",
"\n",
"disp('a president ,a secretary and a treasurer has to be elected in an orga-nisation of 26 members.No person is elected to more than one postion')\n",
"t=26;  //total number of members in the organisation\n",
"j=t*(t-1)*(t-2); \n",
"disp(j,'number of ways to elect the three officers (president,secretary,treasurer')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.3: Factorial_notation.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"disp('To find: factorial of a 6')\n",
"facto2=2*1;\n",
"facto3=3*facto2\n",
"facto4=3*facto3\n",
"facto4=4*facto3\n",
"facto5=5*facto4\n",
"facto6=6*facto5\n",
"\n",
"k=8*7*factorial(6)/factorial(6);\n",
"disp(k,'value of 8!/6! is:')\n",
"j=12*11*10*factorial(9)/factorial(9);\n",
"disp(j,'value of 12!/9! is:')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.4: Binomial_coefficients.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"function [k]=func1(n,r)  //calculating binomial coefficient\n",
"k=factorial(n)/(factorial(r)*factorial(n-r));\n",
"endfunction\n",
"func1(8,2)\n",
"func1(9,4)\n",
"func1(12,5)\n",
"func1(10,3)\n",
"func1(13,1)\n",
" \n",
"p = factorial(10)/(factorial(10-7)*factorial(7));  //calculating 10C7\n",
"q= factorial(10)/(factorial(10-3)*factorial(3))  //calculating 10C3\n",
"disp(p,'value of 10C7 is')\n",
"//10-7=3 so 10C7 can also be computed as 10C3\n",
"//both p and q have same values but second method saves time and space"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.5: Permutations.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"disp('finding the number of three-letter words using only the given six letters(A,B,C,D,E,F) without repetition')\n",
"n=6;  //total number of letters\n",
"l1=n;  //number of ways in which first letter of the word can be chosen\n",
"l2=n-1; //number of ways in which second letter of the word can be chosen\n",
"l3= n-2; //number of ways in which third letter can be chosen\n",
"k=l1*l2*l3;\n",
"disp(k,'number of three-letter words  possible')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.6: Permutations_with_repetitions.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"function [k]= funct1(n,p,q)\n",
"k= factorial(n)/(factorial(p)*factorial(q));\n",
"endfunction\n",
"k=funct1(7,3,2)  //in 'BENZENE' three letters are alike(the three Es) and two are alike (the two Ns)\n",
"disp(k,'The number of seven-letter words that can be formed using letters of the word BENZENE')\n",
"\n",
"disp('a set of 4 indistinguishable red coloured flags, 3 indistinguishable white flags and a blue flag is given')\n",
"j=funct1(8,4,3); \n",
"disp(j,'number of different signals ,each consisting of eight flags')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.7: Combinations.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"disp('four objects are given (a,b,c,d) and three are taken at a time')  \n",
"combinations = factorial(4)/(factorial(4-3)*factorial(3));\n",
"disp(combinations,'number of combinations of the four objects given')\n",
"k=factorial(3);  //number of permutations of objects in a combination\n",
"permutations = combinations*k;\n",
"disp(permutations,'total number of permuatations for the problem')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.8: Combinations.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"function [k]= myfunc(n,r)\n",
"k=factorial(n)/(factorial(n-r)*factorial(r));\n",
"endfunction\n",
"k=myfunc(8,3);  \n",
"disp(k,'the number of committees of three that can be formed out of eight people')\n",
"  \n",
"cows=myfunc(6,3)  //number of ways that a farmer can choose 3 cows out of 6 cows\n",
"pigs=myfunc(5,2)  //number of ways that a farmer can choose 2 pigs out of 5 pigs\n",
"hens=myfunc(8,4)  //number of ways that a farmer can choose 4 hens out of 8 hens\n",
"p=cows*pigs*hens;       \n",
"disp(p,'total number of ways that a farmer can choose all these animals')"
   ]
   }
,
{
		   "cell_type": "markdown",
		   "metadata": {},
		   "source": [
			"## Example 6.9: Combinations_with_repetitions.sci"
		   ]
		  },
  {
"cell_type": "code",
	   "execution_count": null,
	   "metadata": {
	    "collapsed": true
	   },
	   "outputs": [],
"source": [
"//each solution  to the equation can be viewed as a combination of objects\n",
"r=18;  //number of objects \n",
"M=3;   //kinds of object\n",
"m=factorial(r+(M-1))/(factorial(r+(M-1)-(M-1))*factorial(M-1));\n",
"disp(m,'number of non negative integer solutions of the given equation x+y+z=18')"
   ]
   }
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