 Hi, and welcome to the session. Let us discuss the following question. The question says, find the number of four digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of this will be even? Before solving this question, we should first be well versed with theorem 1. The theorem 1 states that the number of permutations in different objects, objects do not repeat, number of different objects, number of objects taken at a time. And this NPR is equal to r factorial. We will use this to solve this question. Begin with the solution. Given five digits r, 4 and 5, we have to find number of by using these digits if no digit is repeated. And also we have to find number of even four digit numbers. You should know that here, 2, 3, 4, 2, 1, are different numbers. The digit numbers with digits is repeated is equal to number of permutations in digits taken 4 at a time. We know that number of permutations of n different objects taken r at a time and objects do not repeat is NPR. Now here n different objects are five different digits and r is 4 and also the digits do not repeat. So by using theorem 1, number of permutations of five different digits taken 4 at a time is equal to 5PR. And as nPR is equal to n factorial upon n minus r factorial, therefore 5PR is equal to 5 factorial upon 5 minus 4 factorial. And this is equal to 5 factorial upon 1 factorial. Now 5 factorial is equal to 5 into 4 into 3 into 2 into 1 factorial. So we have 5 into 4 into 3 into 2 into 1 factorial upon 1 factorial. Canceling 1 factorial from both numerator and denominator we are left with 5 into 4 into 3 into 2. And this is equal to 120. Total number of four digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 and no digits repeated is 120. Now we have to find number of even numbers from this four digit numbers. We know that unit place of even numbers is even. Now the five digits given to us are 1, 2, 3, 4 and 5. Out of these five digits only two that is 2 and 4 are even. So that means we have only two options for filling the unit space. If we fix 2 at the unit space of four digit numbers equal to permutations different digits 3 at a time. And we know that number of permutations of n different objects taken are at a time. And the objects to not repeat is npr. Here n different objects are four different digits and r is 3. So by using theorem 5 permutations of remaining four different digits taken 3 at a time is 4p3. We know that npr is equal to n factorial upon n minus r factorial. Here n is for an r is 3 so 4p3 is equal to 4 factorial upon 4 minus 3 factorial. And this is equal to 4 factorial upon 1 factorial. And this is equal to 4 into 3 into 2 into 1 factorial upon 1 factorial. We can cancel one factorial from both numerator and denominator. So we are left with 4 into 3 into 2 and this is equal to 24. Similarly when we fix 4 at the unit space then number of four digit numbers is equal to number of permutations of remaining four different digits taken 3 at a time that is 4p3. And on simplifying 4 to 3 we get 24. Require even numbers 24 plus 24 and this is equal to 4p8. The required answer is 120. This completes the session.