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I have to find the total number of permutations of four letters that can be selected form the

word "ARRANGEMENT".

Clearly we have 7 different letters so the amount of 4 letter permutations with no repeats is:

7!/3!=840

now for each two letter can form a four letter permutaion with another two different letters

4!/(2!*2!)*6*5

where the first part gives the number of permutations of a four letter word with 2 letters the same. Each of the remaining slots can take one of the other 6 letters and the other by one of the remaining 5 letters.

Now given that there are 4 of these 180*4=720

Finally must look at all the combinations of the double letters to form a four letter permutation:

4!/(2!*2!)*3*4=72

The first part is te number of permutations of a given two letters within a four letter sequence. This is then multiplied by the number of reminaing double letters it may forma permutation with 3. This total is then multiplied by 4 the total number of double letters as any oneof them could form the inital set of permutations.

so I get a total of 72+720+840=1632.

However correct answer is 1596 can someone please explain?

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# Permutations problem

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