If the letters of the word "MISSISSIPPI" are written down at random in a row, the probability that no two S's occur together is:(A) 1/3(B) 7/33(C) 6/13(D) 5/7

If the letters of the word "MISSISSIPPI" are written down at random in

1.	If the letters of the word "MISSISSIPPI" are written down at random in a row, the probability that no two S's occur together is:(A) 1/3(B) 7/33(C) 6/13(D) 5/7
Answer» In the word "MISSISSIPPI", Number of M's = 1 Number of I's = 4Number of S's = 4 Number of P's = 2 Total number of letters = 11 So total number of permutations = 11! / (1! 4! 4! 2!) = 34650 Now, total number of permutations taking all the letters except the S's = 7! / (1! 4! 2!) = 105 After these 7 letters are permuted in a row, a total of 8 places (6 in between positions 2 end positions) are obtained, in which remaining 4 S's can be placed so that no two S's occur together. Number of ways in which 4 S's can be placed in 8 positions is 8C4 = 70. So the number of permutations taking all the letters so that no two S's occur together is = 105 x 70 = 7350 Therefore, the required probability = 7350 / 34650 = 7 / 33