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Re: permut&combi by 50 cent - Monday, 23 July 2012, 10:21 AM

Hi all, There will be 4 cases in total: CASE-1: When there are 2 letters in the row 1& 2 and remaining in row 3 so solving this will give  (2C2*7*6) *(2C2*5*4) * (6C3*3!) = 100800 CASE-2: When there are 2 letters in the row 1& 1 letter in row 2 and remaining in row 3 so solving this will give (2C2*7*6) * (2C1*5) * (6C4*4!)  = 151200 CASE-3: When there 1 letter in the row 1& 2 letters in row 2 and remaining in row 3 so solving this will give (2C1*7) * (2C2*6*5) * (6C4*4!) = 151200 CASE-4: When there 1 letter in the row 1& 1 letter in row 2 and remaining in row 3 so solving this will give 2 {(2C2*7) * (2C2*6)} * (6C5*5!) = 120960.   So the total ways would be 100800+151200+151200+120960 = 524160 Show parent | Reply

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