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InterviewSolution
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In how many ways can the letters of the word PERMUTATIONS bearranged if the(i) words start with P and end with S,(ii) vowels are all together,(iii) there are always 4 letters between P andS? |
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Answer» (i) There are remaining 10 letters in the words starting with P and ending with S in which T occurs twice. No. of such words `= (10!)/(2!) = 1814400`. (ii) If all vowels A,E,I,O,U are together then taking them as on letter, the number of ways of arranging, 8 letters in which T occurs twice `=(8!)/(2!) = (40320)/(2) = 20160` Again, no. of ways of arranging 5 vowels `.^(5)P_(5) = 5! =120` Therefore, total words in which vowels are together `= 20160 xx 120 = 2419200` (iii) The 12 letters of the word PERMUTATIONS can filled at 12 places in the following way: 1 2 3 4 5 6 7 8 9 10 11 12 If there are always four letters between P and S then we can place P at 1,2,3,4,5,6, or 7th place and as a result S can be placed at 6,7,8,9,10,11 or 12th place. So, no of ways to place `P = 7` No. of ways to place `S = 7` Total ways `= 7 +7 14` No. of ways to place remaining 10 letterss at remaining 10 places in which T occur twice. `= (10!)/(2!) = (3628800)/(2) = 1814400` Therefore, required number of words `= 1814400 xx 14` `= 25401600` |
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