Hi, 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?
A,R,N,E occur twice and G,M,T occur once. 1/ No. of arrangement when all letters different: =7x6x5x4=840 2/ No. of arrangement when 2 are of one kind and others different:= (no. of ways to select the letter to be repeated) x (no. of ways to select the letters not to be repeated) x (no. of ways to arrange them) 4C1 x 6C2 x 4!/2! = 4 x (6x5/2) x 4!/2! =720 [6C2= 6 combination 2] 3/ No. of arrangement when two letters used (each repeated twice):= 4C2 x 4!/(2!x2!) = 36 840+720+36=1596