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You would only need to divide by 2! for the S's if you first multiplied by 2! to arrange themDreamerish*~ said:EDIT: Actually, 7!/4, since there are two S's in MISS. I'm still not sure.
Um. No.jake2.0 said:i did another way, consider there to be to spaces so _ _ _ _ _ _ _ _ _ _ . lets say that MISS appears at the start so for the first _ there is 1 possible solution (couldn't find right word). for the second _ u choose one of the 3 I's so there are 3 possible solutions. for the third _ u choose one of the 2 S's so u have 2 possible solutions. for the fourth _ u choose the remaining S so u have 1 possible solution. for the rest of the _'s its just 6!
so total ways = 1x3x2x1x6!
= 4320
what 2 are u saying are the same, the 2 S's in MISS or the 2 I's left over?KFunk said:If the word MISS is to appear then you can't really have MSIS or SMIS so treat [MISS] as a single unit in itself (no arrangements required... if you switch the S's nothing has changed). You're then left with 6 letters, 2 of which are identical. You can treat [MISS] as another individual letter unit which leads you to dreamerish's first result:
7!/2! (the number of permutations of 7 objects with 2 the same)
LOL.KFunk said:- miss
- miss
^ spot the difference
After you take MISS out, there are 6 letters left over, but when you work out how many combinations there are, you include MISS as one group, so there are 7 "letters".dz said:isnt it 6!
ok that depends on permutations or combinationsKFunk said:- miss
- miss
^ spot the difference
umm... no it doesnt, it applies for both lolrnitya_25 said:ok that depends on permutations or combinations