# 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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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

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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