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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > In how many ways can the letters of the PERMUTATIONS...
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In how many ways can the letters of the PERMUTATIONS be arranged if there are always 4 letters between P and S ?

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I figured out the solution however it is not a simple permutation question. This includes both permutation as well as combination. There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice. Now first we need to see how many ways we can make word with 4 letter between P and S. Except...
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I figured out the solution however it is not a simple permutation question. This includes both permutation as well as combination. There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice. Now first we need to see how many ways we can make word with 4 letter between P and S. Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210 Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S. So total way = 210*2 = 420 The selected 4 letters can be rotated between P and S in = 4! ways So total ways = 420 * 4! Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter. Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways. So total number of ways = 7! * 420 * 4! Now since letter T was repeated twice, we should divide the above result by 2!. So Total number of ways = 7! * 420 * 4! / 2! = 25401600 read less
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This is a 12 letter word. Let each position of the word be numbered. So there are 12 positions to fill in. _ _ _ _ _ _ _ _ _ _ _ _ if P is in position 1 then 'S' will be in number 6, position 2,3,4,5 will be occupied by some other four letters. Let us call this pair (1,6) Similarly for position...
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This is a 12 letter word. Let each position of the word be numbered. So there are 12 positions to fill in. _ _ _ _ _ _ _ _ _ _ _ _ if P is in position 1 then 'S' will be in number 6, position 2,3,4,5 will be occupied by some other four letters. Let us call this pair (1,6) Similarly for position of 'P' at 2 ,3 4, 5 ,6 and 7. This gives 7 cases. If we reverse order between p and s we get another 7 cases Total 14 cases. Leaving out p and s there are 10 letters with t coming twice. These can take any of those 10 places left out by gaps in 10!/2! ways. So total 14* 10!/2! = 7 * 10*9*8*7*6*5*4*3*2 = 25401600 read less
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Ans: 7*2!*(10!/2!) = 7*10! P _ _ _ _ S _ _ _ _ _ _(One way to place P and S P and S can be arranged with a 4 letter difference in between them in 7*2! ways. Remaining 10 letters (E R M U A I O N TT) can be arranged in 10!/2! ways.
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There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice. 1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210; 2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2; 3. Permutation of the 4 letters between P and S = 4! =24; 4....
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There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice. 1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210; 2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2; 3. Permutation of the 4 letters between P and S = 4! =24; 4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040; 5. We should divide multiplication of the above 4 numbers by 2! as there is repeated T. Hence: (10c4 x 2! x 4! x 7!) / 2! = 25,401,600 read less
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( 7! * 420 * 4! )/ 2 = 25401600
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Don't want to over express myself, just join and feel the experience.

In the 12 letter word there are 14 different places where P and S can be separated by 4 letters. Now the remaining 10 letters can be arranged in 10!/2! ways as T repeats two times . So total no. of ways arrangement with the given condition is 14*10!/2! = 25401600
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