 Hi, and welcome to this session. Let us discuss upon a question. The question says, how many words with or without meaning can be formed using all the letters of the word, equation, using each letter exactly once? Before solving this question, we should first be able to most with theorem one. The theorem one states that the number of commutations of different objects taken are at a time, jets do not repeat, is the number of different objects, or is the number of objects taken at a time. 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. Begin with the solution. Letters of word, equation are E, Q, U, A, without meaning that can be formed using all these letters when each letter is used exactly once. Right, each letter is different, and each letter has to be used only once. No repetition is allowed. Therefore, number by using equation is equal to number of commutations of A different letters taken all one. We know that number of commutations of indifferent objects taken are at a time at the objects do not repeat, is NPR. Here, N different objects are A different letters, and R is also A, since all the letters has to be used. So by using theorem one, number of commutations of A different letters taken all at a time is equal to 8 P8. We know that NPR is equal to N factorial upon N minus R factorial. N is equal to 8, and R is also equal to 8. So by using this formula, 8 P8 is equal to 8 factorial upon 8 minus 8 factorial. And this is equal to 8 factorial upon 0 factorial. 8 factorial is equal to 8 into 7 into 6 into 5 into 4 into 3 into 2 into 1 into 0 factorial upon 0 factorial. You can cancel 0 factorial from both numerator and denominator. So we are left with 8 into 7 into 6 into 5 into 4 into 3 into 2 into 1. And on simplifying it, we get 4, 0, 3, 2, 0. Hence, our required answer is 40,320. This now meets the session. Bye and take care.