Submitted by Some.Username on Thu, 08/09/2018 - 10:47
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The Hints and Partial Solutions link just links back to the questions so there aren't any answers for Assignment 24.
Submitted by Some.Username on Thu, 08/09/2018 - 10:47
The Hints and Partial Solutions link just links back to the questions so there aren't any answers for Assignment 24.
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Thanks!
Should be fixed now. Thanks for bringing this to our attention!
Assignment 6 (2 ii)
For the distinct ways Ebenezer can rearrange his name. Shouldn't the solution be 8!/5!. Why is it 8!/4! (Mentioned in the hints and partial solutions?
Ebenezer
If you treat all the "e's" as different letters (i.e. consider the number of permutations of $e_1, b, e_2, n, e_3, z, e_4, r $ there are $8!$ different ways of arranging this. However there are $4!$ ways of arranging $e_1, e_2, e_3, e_4$, so these permutations occur in "groups" of $4!$ permutations which are in fact the same permutation. Hence the answer is $\frac{8!}{4!}$.