In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S? The answer is 25401600. Can anybody explain me how?

First, determine the number of ways to fill out the PXXXXS unit. This unit is 6 letters long: _ _ _ _ _ _. First determine the possibilities for the first and last letters. We have 2 choices for the first letter and 1 for the last letter: 2 _ _ _ _ 1. Now, there are ten remaining letters that could go in the blanks, so 2*10*9*8*7*1 total ways to determine this unit. Now, treat the unit as one and arrange it with the remaining 6 letters which could be done in 7! ways. Multiply these together to get 2*10*9*8*7*1*7!. Finally the letter T appears twice in PERMUTATIONS, so divide all of this by 2 to get 10*9*8*7*7! or [spoiler]7*10!=25401600[/spoiler]

PERMUTATIONS P/S _ _ _ _ P/S _ _ _ _ _ _ => 10!/2 * 2 _ P/S _ _ _ _ P/S _ _ _ _ _ => 10!/2 * 2 _ _ P/S _ _ _ _ P/S _ _ _ _ => 10!/2 * 2 _ _ _ P/S _ _ _ _ P/S _ _ _ => 10!/2 * 2 _ _ _ _ P/S _ _ _ _ P/S _ _ => 10!/2 * 2 _ _ _ _ _ P/S _ _ _ _ P/S _ => 10!/2 * 2 _ _ _ _ _ _ P/S _ _ _ _ P/S => 10!/2 * 2 Total no of ways = 7 * 10!/2 * 2 = 25401600

dinaroneo wrote:In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S? The answer is 25401600. Can anybody explain me how? No. of letters in the word PERMUTATIONS = 12 If P is placed at 1st blank and S at the 6th place, P _ _ _ _ S _ _ _ _ _ _ Similarly, if P is placed at 2nd blank and S at the 7th place, _ P _ _ _ _ S _ _ _ _ _ This way there are 7 ways of doing so. Also P and S can be interchanged, viz. S at the 1st place and P at the 6th place. So, no. of ways of placing P and S = 7 * 2 = 14 Corresponding to these 14 ways, the other 10 digits can be placed in 10 places as 10!/2! = 1814400 Therefore, required no. of ways = 14 * 1814400 = 25401600

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