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2c2*(4 c 1+4 c 2+4 c 3+4 c 4) = 15.
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I will give it a hand,
6 letters , 2 same, can be arranged in 6!/2 different ways. When 2 are together in 5! different ways. Then the ans should be 6!/2-5!=240 ways when there wil be AT LEAST one letter in between.

Prof, i think that you select the letters out of the remaining 4 different but shouldn't the permutations of these selections be considered? :?
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BG
I will give it a hand,
6 letters , 2 same, can be arranged in 6!/2 different ways. When 2 are together in 5! different ways. Then the ans should be 6!/2-5!=240 ways when there wil be AT LEAST one letter in between.

Prof, i think that you select the letters out of the remaining 4 different but shouldn't the permutations of these selections be considered? :?


We should use combinations and not permutations. This is because which letter is arranged first has no meaning to the solution. We're only interested in what is inside a group of letters selected.



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