 Hello and welcome to the session. In this session we are going to discuss the following question which says that what is the probability that the numbers formed with the digits 1, 2, 3, 4, 5, 6, 7, 8 are such that the odd digits always occupy the odd places and even digits always occupy the even places and digits cannot be repeated. The number of odd permutations of n distinct things taken all of the time is given by n factorial. We also have fundamental principle of multiplication takes that if there are two jobs to be done such that the first job can be done in n number of ways, the second job can be done in n number of ways, then both the jobs together can be done in n into n number of ways. With this key idea let us proceed with the solution. We need to find the probability that the numbers formed with the digits 1, 2, 3, 4, 5, 6, 7, 8 such that the odd digits always occupy the odd places and the even digits always occupy the even places. Also the digits cannot be repeated. We are given 8 digits so we have 8 places to fill in. There are 4 odd places that is 1st, 3rd, 5th and 7th and 4 even places that is 2nd, 4th, 6th and 8th. So 4 odd digits can occupy odd places such that now 1st place can be occupied by any of the 4 digits. So there are 4 options for this place. Now 3rd place can be occupied by remaining 3 odd digits. There are 3 options for this place similarly 2 options and the 7th place will have 1 option. Therefore the number of ways in which 4 odd digits are occupied 4 odd places is given by 4 into 3 into 2 into 1 that is equal to 4 factorial. Similarly there are 4 even places that is 2nd, 4th, 6th and 8th and 4 even digits given these 4 even places. Therefore the number of ways 4 even digits can occupy 4 even places is given by 4 factorial. We need to find the probability that the numbers formed with these 8 digits are such that the odd digits always occupy the odd places and even digits always occupy the even places. For this we have the fundamental principle of multiplication which states that if there are 2 jobs to be done such that the first job can be done in n number of ways and second job can be done in n number of ways then both the jobs together can be done in n into n number of ways. Here our first job will be the number of ways in which 4 odd digits can occupy the 4 odd places which is given by 4 factorial and the second job would be the number of ways in which the 4 even digits can occupy the 4 even places which is given by 4 factorial. Therefore by using the fundamental principle of multiplication we can say that the number of favorable outcomes be equal to 4 factorial into 4 factorial and from the key idea we know that the number of all permutations of n distinct things taken all at a time is given by n factorial. Therefore the number of ways in which we occupy places is given by 8 factorial. Total number of outcomes will be factorial. Therefore the required probability is given by number of favorable outcomes upon total number of outcomes that is 4 factorial into 4 factorial upon 8 factorial which is equal to 4 factorial into 4 factorial and 8 factorial can be written as 8 into 7 into 6 into 5 into 4 factorial which is equal to 4 factorial goes with 4 factorial. So we are left with 4 factorial which can be written as 4 into 3 into 2 into 1 upon 8 into 7 into 6 into 5 which is equal to 1 upon 7 into 2 into 5 which is equal to 70. Therefore the required probability is equal to 1 upon 70 which is the final answer. This completes our session but we will enjoy this session.