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In how many ways can the letters of the word ASSASSINATION be arranged so that all the S’s are together ?

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Solution: Group all 'S' together and call it X 
remaining letters have 3A, 2I, 2N, one T and one O 
so total letters = 10 
out of this 3! terms will have indistinguishable 'A' 
2! indistinguishable 'I' and same for N 
So total = 10!/(3!*2!*2!) 
=151200

+5 votes
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The word "ASSASSINATION" has 4S , 3A , 2I , 2N , T , O , 4S are together. this is considered as one block as 1 letter. we have 3A , 2I, 2N , 4S , T , O 

Therefore , Number of words = Number of Permutations of 3A , 2I , 2N , 4S , T , O = 10! / 3!2!2! 

= 10X9X8X7X6X5X4X3X2X1 / (3X2X1) X (2X1) X (2X1)

= 10X9X8X7X5X3X2 = 151200

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