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Class 12 Tuition > Learn Class XI-XII Tuition (PUC) > In how many ways can the letters of the word "ASSASSINATION"...
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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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Math Educator for Std.11th ,12th , Engineering Entrance and Degree Level with 11+ Years Experience

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)...
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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 read less
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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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10!/(3!*2!*2!)
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Tutor - Maths and computer science

The word ASSASSINATION has 4S,3A,2I,2N,T,O,4S are together. This is considered as one block as 1 letter Now we have 3A,2I,2N,4S,T,O ?10!3!2!2!=10×9×8×7×6×5×4×3×2×1(3×2×1)×(2×1)×(2×1) ?10×9×8×7×5×3×2 ?151200
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B.Tech IIT Delhi,maths lover who loves to teach

akshay you need to put much efforts in maths bro, better start reading any book, do solved examples. Putting up a question on this website wont help
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Native English Speaker with 32 years experience in teaching English and communication skills.

907200
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SSS are to be together. ie consider SSSS as a single free unit. remaining free units= (13 - 4) Total number of free units = (13 -4) +1= 10 free units 10 free units can be arranged in 10 ! ways Internal arrangement of SSSS together = 4 ! ways Total number of arrangements = 10! x 4 ! = 87091200...
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SSS are to be together. ie consider SSSS as a single free unit. remaining free units= (13 - 4) Total number of free units = (13 -4) +1= 10 free units 10 free units can be arranged in 10 ! ways Internal arrangement of SSSS together = 4 ! ways Total number of arrangements = 10! x 4 ! = 87091200 ways But Identical permutations are= 87091200/(4! x 3! x 2! x 2!) =151200 ways read less
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Mathematics for JEE Mains/Advanced, XI & XII (All Boards)

10! / 3!2!2!
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SME in Mathematics & Statistics

In the word "ASSASSINATION" A, S, I and N repeated 3 times, 4 times, 2 times and 2 times respectively. Since 4S must be together, SSSS can be treated as one entity. Hence, total number of arrangement = 10! / (3! x 2! x 2!) = 151200
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151200
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