 Hi and welcome to the session. Let us discuss the voting question. The question says in how many ways can the letters of the word for mutations be arranged if the first part is if the words start with p and end with s. Let's now begin with the solution. In this part we have to find number of words starting with p and ending with s in the word permutations. So we will fix p at the first place at the last place. As we have fixed p at the first place and s at the last place therefore we are left with 10 letters which are e, r, e appears two times and the rest are all different. Number of mutations, different letters, letters are t's and rest are all different equal to 10 factorial upon 2 factorial by theorem 4 which states that the number of four mutations of n objects where p1 objects are of one kind, p2 are of second kind, pk are of k kind and the rest if any are different kind is n factorial upon p1 factorial into p2 factorial and so on up to pk factorial. Now 10 factorial upon 2 factorial is equal to 10 into 9 into 8 into 7 into 6 to 5 into 4 into 3 into 2 factorial upon 2 factorial. Now 2 factorial gets cancelled out from both different denominators so we are now left with 10 into 9 into 8 into 7 into 6 into 5 into 4 into 3 and this is equal to 181 double 4 0 0. Hence I require the answer is 181 4 4 0 0. This completes the first part let's now move on to the second part. Second part is vowels are all together. Now there are 12 letters in the word for mutations and we have to find number of words where vowels are all together. Now the vowels in the word for mutations are e, u, a, i and o. Now we take all vowels that is e, u, a, i, o as one unit. Now this unit will occupy one place our word is made of e, u, a, i, o and rest of the letters that is so now we have 8 letters e appears 2 times rest are all different. Number of for mutations of 8 different letters and the rest are different is equal to 8 factorial upon 2 factorial by theorem 4. For vowels can have number of for mutations of 5 different letters taken 5 at a time. So number of for mutations of 5 different letters taken 5 at a time the letters do not repeat 5. Now by multiplication principle required number of for mutations is equal to 8 factorial upon 2 factorial into 5 p5 and this is equal to 8 factorial upon 2 factorial into 5 factorial and this is equal to 8 into 7 into 6 into 5 into 4 into 3 into 2 factorial upon 2 factorial and 5 factorial is equal to 5 into 4 into 3 into 2 into 1. Now 2 factorial gets cancelled out and this is equal to 2 0 1 6 0 into 1 20 and this is equal to 2 4 1 9 2 0 0. Hence our required answer is 2 4 1 9 2 0 0. This completes the second part let's now move on to the last part. Third part is there are always 4 letters between p and s. In this part we have to find number of words where there are always 4 letters between p and s of the word permutations. Now the 12 letters of the word permutations are p e a have to find number of words where there are always 4 letters between p and s. The position 1 at position 6. We have 12 letters so we will have 12 positions that is 1 2 3 4 5 6 7 8 9 10 11 and 12. Now we have fixed p at the first position and s at the sixth position so that there are 4 letters in between p and s. Now after fixing p and s the remaining letters are there are 10 letters which are left after fixing p at the first position and s at the sixth position and out of these 10 letters t appears 2 times and the rest are all different. Number of permutations between letters where 2 letters and rest are different is equal to 10 factorial upon 2 factorial by theorem 4. We can also change places thus we will have 2 factorial permutations till number of permutations at position 1 position 6 is equal to 10 factorial upon 2 factorial into 2 factorial. There are other situations also where there are 4 letters between p and s. So now we will list all these situations. Now these are the 7 situations where there are always 4 letters are between p and s. Now by multiplication principle the number of permutations of the word permutations is equal to 7 into 10 factorial upon 2 factorial into 2 factorial and this is equal to 7 into 10 into 9 into 8 into 7 into 6 into 5 into 4 into 3 into 2 into 1 and this is equal to 7 into 3 6 2 8 8 0 0 and this is equal to 2 5 4 0 1 6 0 0 hence our required answer is 2 5 4 0 1 6 0 0. This is our required answer. So this completes the session. Bye and take care.