Three letters from the word freedom are selected and arranged in a row. how many different arrangements of the three letters are obtained, given that repetition of letters is allowed?

You can pick 3 letters from the group of 7 letters by combination. C(7,3)=35. Then you you can arrange them in 3!=6 ways. For example you can pick 3 letters for the first place, there left 2 letters, then pick one of the 2 letters for the second place and last letter goes to the third place. In the end you should multiply 35 with 6. Think each choice of picking 3 letters and arrangements as different possibilities. The result should be 35*6=210. Hope it works :)

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