 Hello and welcome to the session. In this session first we are going to discuss fundamental principle of counting. According to this we have that if an event can occur in n different ways following which another event can occur in n different ways then the total number of occurrence of events in the given order is m into n. This fundamental principle can be generalized for any finite number of events like for 3 events if we have that if an event can occur in m different ways following which another event which occurs in n different ways following which a third event can occur in p different ways then the total number of occurrence to the events in the given order is m multiplied by n multiplied by p. Consider a cinema hall in which we have 3 entrance doors and 2 exit doors. Let's see in how many ways a person can enter the hall and then come out. Clearly since there are 3 entrance halls so person can enter the hall through any of the 3 entrance doors so we say that there are 3 ways of entering the hall. Then after entering the hall the person can come out through any of the 2 exit doors and so we can say that there are 2 ways of coming out of the hall. Thus the total number of ways in which a person can enter and then come out of the hall is equal to 3 multiplied by 2 and that is equal to 6. Next we have permutations basically a permutation is an arrangement in a definite order of a number of objects taken some or all at a time. Now we have the number of permutations of n different objects taken r at a time where we have r is greater than 0 and less than equal to n and the objects do not repeat is n multiplied by n minus 1 multiplied by n minus 2 and so on up to n minus r plus 1 which is generally denoted by npr. The formula for npr is given as n factorial upon n minus r factorial where we have r is greater than equal to 0 and less than equal to n and we know the factorial notation is given by n factorial is equal to 1 multiplied by 2 multiplied by 3 multiplied by 4 and so on up to n minus 1 multiplied by n or we can also say that n factorial is equal to n multiplied by n minus 1 factorial. Let's calculate 12 factorial here we have n is 12 and r is 4 so we get 12 p4 is equal to 12 factorial upon 12 minus 4 that is 8 factorial this is equal to 12 into 11 into 10 into 9 into 8 factorial upon 8 factorial. Now this 8 factorial is cancelled with this 8 factorial and we are left with this which is equal to 11 double 8 0 so we have 12 p4 is equal to 11 double 8 0. Now we have another important result according to which we have the number of permutations of n different objects taken r at a time where repetition is allowed is given by n to the power r. Consider the word office in this we have six letters so here we have n is equal to 6. Now let's try and find out the number of three letter words which can be formed by the letters of this word when repetition is allowed this would be given by 6 to the power 3 here n is 6 and 3 is r and this is equal to 216 that is we can form 216 words from the letters of the word office when repetition is allowed. Now we discuss permutations when all the objects are not distinct objects the result in this case goes like the number of permutations of n objects where p objects are of the same kind rest r or different is equal to n factorial upon p factorial. In fact we have a more general result which says number of permutations of n objects where p1 objects are of one kind p2 are of second kind and so on pk are of kth kind and the rest if any are of different kind is given by n factorial upon p1 factorial multiplied by p2 factorial and so on up to pk factorial. Let's try and find out the number of permutations of the letters of the word office. Now there are six letters in this word so we say that there are six objects that is we can say that n is equal to 6 here. Now as you can see two objects are of same kind and rest are all different so we say required number of permutations is equal to n factorial that is 6 factorial upon 1 factorial for the letter o then 2 factorial for the letter f which is occurring two times in the word office then into 1 factorial for the letter i again multiplied by 1 factorial for the letter c which is again multiplied by 1 factorial for the letter e. Now this is equal to 6 multiplied by 5 multiplied by 4 multiplied by 3 multiplied by 2 factorial upon 2 factorial now 2 factorial 2 factorial gets cancelled and we are left with this which is equal to 360. So there are 360 permutations of the letters of the word office. This completes the session hope you have understood the concept of permutations.