Tom K. answered 04/25/21
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There are 10 letters, but not 10 different ones.
2 e's
1 l
1 r
2 s's
3 t's
1 u
Thus, there are 10!/2!1!1!2!3!1! different orders.
This equals 3628800/24 = 151200
William A.
My problem is with the space, Should I consider them like two distinct word Test 2 t’s 1e 1s 4!/2!1!1! That’s 12 result R = 1; E = 1; S = 1; U = 1; L = 1; T = 1; 6! = 720 For each arrangement in test we have 720 arrangement in result So I multiply it 720 * 12 = 8640 Like if we have “abc ab” The answer will be abc ab abc ba acb ab acb ba bac ab bac ba bca ab bac ba cab ab cab ba cba ab cba ba So it will be 12 or if I do it like you the answer will be 3004/25/21