Step 1: What is permutation

DEFINITION: Permutation: Each of several possible ways in which a set

or number of things can be ordered or arranged.

Number of permutations of the letters of the word which has n letters is n! If some of letters is doubled Number of permutations is \frac{n!}{2} , 2 letters are doubled Number of permutations is \frac{n!}{2*2} …

Step 2: Analyse the given question and set up a formula

FORMULA: Therefore, required number of possible permutations where neither ‘HIN’ nor ‘DUS’ nor ‘TAN’ will be together = The total number of permutations - [(number of permutations which HIN come as block, +TAN come as block + DUS comes as a block) - (number of permutations when both HIN and TAN come as block + TAN and DUS come as block + DUS and TAN comes as blocks) + (permutations of three blocks come as blocks)]

Step 3: Calculate the total number of permutations.

EXAMPLE: In the word ‘HINDUSTAN'; number of letters = 9 out of which N

comes two times. So, the total ways of possible

permutations = \frac{9!}{2!} = 181440.

Step 4: Find the number of permutations when HIN come as block, TAN come as block and DUS comes as a block.

EXAMPLE: Number of permutations in which HIN comes as a block = 7!

7! = 7*(7-1)*(7-2)*( 7- 3)*(7 - 4)*(7-5)*(7-6) = 5040

Number of permutations in which TAN comes as a block = 7! = 5040

Number of permutations in which DUS comes as a block = \frac{7!}{2} = 2520

Step 5: Determine the permutations come as a block like (‘HIN’ and ‘DUS, TAN’ and ‘DUS’ and TAN’ and ‘HIN’ )

EXAMPLE: The total number of permutations where ‘HIN’ and ‘DUS’ will

be together = 5! = 120

The total number of permutations where ‘TAN’ and ‘DUS’ will be

together = 5! = 120

The total number of permutations where ‘TAN’ and ‘HIN’ will be

together = 5! = 120

Step 6: Find the total number of permutations where ‘HIN’, ‘DUS’ and ‘TAN’ together

EXAMPLE: The total number of permutations where ‘HIN’, ‘DUS’ and ‘TAN’

will be together = 3! = 6

Step 7: Find the required value by using the formula in step 2.

Therefore, required number of possible permutations where neither ‘HIN’ nor ‘DUS’ nor ‘TAN’ will be together

= \frac{362880}{2} - [(5040+2520+5040- 3*(120) + 6)]

= 181440 - [(5040+2520+5040- 3*(120) + 6)]

= 169194.