 Hi and welcome to the session. I'm Kanika and I'm going to help you to solve the following question. The question says how many three-digit even numbers can be made using the digit 1, 2, 3, 4, 6, 7 if no digit is repeated. Before solving this question you should know the theorem 1. Theorem 1 states that the number of commutations in different objects taken r at a time objects not repeat to pr. And this npr is equal to n factorial upon n minus r factorial. We will use this theorem as key idea to solve this question. Next now we begin with the solution. Now given six digits r and seven number of three-digit even numbers can be formed by using these digits if no digit is repeated. You should know that here of digit 2, 1 are different numbers. Since we have to find only three-digit even numbers so this clearly implies that the unit base of three-digit number must be even and we are given six digits like this 1, 2, 3, 4, 6 out of these six digits 2, 4 and 6 are even so we have three options for filling the unit's place. Next at the unit base then the number of three-digit numbers equal to commutations different digits and number of permutations of n different objects taken r at a time and the objects do not repeat is npr. Now here n different objects are five different digits permutations of remaining five different digits taken two at a time is equal to five x4 at the unit's place then the number of three-digit numbers ending with four is equal to permutations of remaining five different digits taken two at a time require number of digit numbers. We can say that digit numbers ending with equal to permutations of remaining five different digits taken two at a time and this is equal to five p2 three-digit numbers number of three-digit even numbers is equal to sum of the permutations of numbers ending with two four or six that is five p2 plus five p2 plus five p2 and this is equal to three into five p2 pr is equal to n factorial upon n minus r factorial two is equal to five factorial minus two factorial and this is equal to three into five factorial is equal to five into four into three factorial upon five minus two factorial is three factorial cancelling three factorial from both numerator and denominator we are left with three into five into four and this is equal to 60 so the required number of three digit even numbers that can be formed with one two three four six and seven when no digit is repeated is 60 this is our required answer so this completes the session bye and take care